1. 程式人生 > 實用技巧 >《演算法筆記》6. 連結串列相關面試題總結

《演算法筆記》6. 連結串列相關面試題總結

目錄

1 連結串列問題

面試時連結串列解題的方法論

對於筆試,不用太在乎空間複雜度,一切為了時間複雜度

對於面試,時間複雜度依然放在第一位,但是一定要找到空間最省的方法

1.1 連結串列面試常用資料結構和技巧

1、 使用容器(雜湊表,陣列等)

2、 快慢指標

1.1.1 快慢指標問題

1、 輸入連結串列頭結點,奇數長度返回中點,偶數長度返回上中點

1 3 5 2 7 返回 5;1 3 2 7 返回 3

2、 輸入連結串列頭結點,奇數長度返回中點,偶數長度返回中下點

1 3 5 2 7 返回 5;1 3 2 7 返回 2

3、 輸入連結串列頭結點,奇數長度返回中點前一個,偶數長度返回上中點前一個

1 3 5 2 7 返回 3;1 3 2 7 返回 1

4、 輸入連結串列頭結點,奇數長度返回中點前一個,偶數長度返回下中點前一個

1 3 5 2 7 返回 3;1 3 2 7 返回 3

package class06;

import java.util.ArrayList;

public class Code01_LinkedListMid {

	public static class Node {
		public int value;
		public Node next;

		public Node(int v) {
			value = v;
		}
	}

	// head 頭
	// 1、奇數長度返回中點,偶數長度返回上中點
	public static Node midOrUpMidNode(Node head) {
	    // 沒有點,有一個點,有兩個點的時候都是返回頭結點
		if (head == null || head.next == null || head.next.next == null) {
			return head;
		}
		// 連結串列有3個點或以上
		// 快慢指標,快指標一次走兩步,慢指標一次走一步
		// 快指標走完,慢指標在中點位置
		Node slow = head.next;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

    // 2、奇數長度返回中點,偶數長度返回中下點
	public static Node midOrDownMidNode(Node head) {
		if (head == null || head.next == null) {
			return head;
		}
		Node slow = head.next;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

    // 3、奇數長度返回中點前一個,偶數長度返回上中點前一個
	public static Node midOrUpMidPreNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node slow = head;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

    // 4、奇數長度返回中點前一個,偶數長度返回下中點前一個
	public static Node midOrDownMidPreNode(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		if (head.next.next == null) {
			return head;
		}
		Node slow = head;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

    // 筆試可以用這種複雜點的方法,空間複雜度比上面快慢指標要高
	public static Node right1(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 1) / 2);
	}

	public static Node right2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get(arr.size() / 2);
	}

	public static Node right3(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 3) / 2);
	}

	public static Node right4(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 2) / 2);
	}

	public static void main(String[] args) {
		Node test = null;
		test = new Node(0);
		test.next = new Node(1);
		test.next.next = new Node(2);
		test.next.next.next = new Node(3);
		test.next.next.next.next = new Node(4);
		test.next.next.next.next.next = new Node(5);
		test.next.next.next.next.next.next = new Node(6);
		test.next.next.next.next.next.next.next = new Node(7);
		test.next.next.next.next.next.next.next.next = new Node(8);

		Node ans1 = null;
		Node ans2 = null;

		ans1 = midOrUpMidNode(test);
		ans2 = right1(test);
		System.out.println(ans1 != null ? ans1.value : "無");
		System.out.println(ans2 != null ? ans2.value : "無");

		ans1 = midOrDownMidNode(test);
		ans2 = right2(test);
		System.out.println(ans1 != null ? ans1.value : "無");
		System.out.println(ans2 != null ? ans2.value : "無");

		ans1 = midOrUpMidPreNode(test);
		ans2 = right3(test);
		System.out.println(ans1 != null ? ans1.value : "無");
		System.out.println(ans2 != null ? ans2.value : "無");

		ans1 = midOrDownMidPreNode(test);
		ans2 = right4(test);
		System.out.println(ans1 != null ? ans1.value : "無");
		System.out.println(ans2 != null ? ans2.value : "無");

	}

}

1.1.2 面試題一:判斷迴文結構

給定一個單鏈表的頭結點head,請判斷該連結串列是否為迴文結構。迴文就是正著輸出和反著輸出結果一樣

  1. 棧的方法特別簡單(筆試用)

筆試思路,以此把該連結串列放入棧中。再遍歷該連結串列和棧中彈出的數比對,只要有不一樣,就不是迴文

  1. 改原連結串列的方法需要注意邊界問題(面試用)

快慢指標解法:用快慢指標定位到中點的位置,奇數就是定位到唯一的中點,偶數定位到上中點。然後把中點右半部分加入棧中去,那麼棧中存的是右半部分的逆序。接著從頭遍歷連結串列,棧中有多少個元素,我們就比較多少步,如果有對不上就不是迴文

快慢指標最優解,不使用容器結構(stack),O(1):同樣的找到中點位置,把右半部分指標回指到中點。接著指標1從L位置,指標2從R位置,往中間遍歷。,每步比對,如果有不一樣,則不是迴文。返回答案之前,把中點右邊的指標調整回來

package class06;

import java.util.Stack;

public class Code02_IsPalindromeList {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	// need n extra space
	public static boolean isPalindrome1(Node head) {
	    // 依次進棧
		Stack<Node> stack = new Stack<Node>();
		Node cur = head;
		while (cur != null) {
			stack.push(cur);
			cur = cur.next;
		}
		// 每個元素和棧中比較
		while (head != null) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need n/2 extra space
	// 中點右側進棧
	public static boolean isPalindrome2(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node right = head.next;
		Node cur = head;
		while (cur.next != null && cur.next.next != null) {
			right = right.next;
			cur = cur.next.next;
		}
		Stack<Node> stack = new Stack<Node>();
		while (right != null) {
			stack.push(right);
			right = right.next;
		}
		while (!stack.isEmpty()) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need O(1) extra space
	// 不使用容器(stack)的方法
	public static boolean isPalindrome3(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		// 慢指標
		Node n1 = head;
		// 快指標
		Node n2 = head;
		while (n2.next != null && n2.next.next != null) { // find mid node
			n1 = n1.next; // n1 -> mid
			n2 = n2.next.next; // n2 -> end
		}
		// n1 中點
		
		
		n2 = n1.next; // n2 -> right part first node
		n1.next = null; // mid.next -> null
		Node n3 = null;
		// 右半部逆序指向中點
		while (n2 != null) { // right part convert
			n3 = n2.next; // n3 -> save next node
			n2.next = n1; // next of right node convert
			n1 = n2; // n1 move
			n2 = n3; // n2 move
		}
		// 引入n3記錄最後的位置,之後把右半部再逆序回原來的次序
		n3 = n1; // n3 -> save last node
		n2 = head;// n2 -> left first node
		boolean res = true;
		while (n1 != null && n2 != null) { // check palindrome
			if (n1.value != n2.value) {
				res = false;
				break;
			}
			n1 = n1.next; // left to mid
			n2 = n2.next; // right to mid
		}
		n1 = n3.next;
		n3.next = null;
		// 把右半部分再逆序回來
		while (n1 != null) { // recover list
			n2 = n1.next;
			n1.next = n3;
			n3 = n1;
			n1 = n2;
		}
		return res;
	}

	public static void printLinkedList(Node node) {
		System.out.print("Linked List: ");
		while (node != null) {
			System.out.print(node.value + " ");
			node = node.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {

		Node head = null;
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(2);
		head.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(2);
		head.next.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

	}

}

1.1.3 面試題二:按值劃分單鏈表

將單鏈表按某值劃分成左邊小,中間相等,右邊大的形式

  1. 把連結串列放入數組裡,在陣列上做partion(筆試用)

[3, 2, 4, 7, 0, 2, 1]選擇2劃分,基於2對陣列作partion

  1. 分成小、中、大三部分。再把各個部分之間串起來(面試用)

藉助6個引用變數,不需要容器O(N),且能保證穩定性。小於區域的頭引用,小於區域的尾引用,等於區域的頭引用,等於區域的尾引用,大於區域的頭引用,大於區域的尾引用。依次對比給定的值加入到這三個區域,之後串聯起來

package class06;

public class Code03_SmallerEqualBigger {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

    // 方法1
	public static Node listPartition1(Node head, int pivot) {
		if (head == null) {
			return head;
		}
		Node cur = head;
		int i = 0;
		while (cur != null) {
			i++;
			cur = cur.next;
		}
		Node[] nodeArr = new Node[i];
		i = 0;
		cur = head;
		for (i = 0; i != nodeArr.length; i++) {
			nodeArr[i] = cur;
			cur = cur.next;
		}
		arrPartition(nodeArr, pivot);
		for (i = 1; i != nodeArr.length; i++) {
			nodeArr[i - 1].next = nodeArr[i];
		}
		nodeArr[i - 1].next = null;
		// 返回頭結點
		return nodeArr[0];
	}

	public static void arrPartition(Node[] nodeArr, int pivot) {
		int small = -1;
		int big = nodeArr.length;
		int index = 0;
		while (index != big) {
			if (nodeArr[index].value < pivot) {
				swap(nodeArr, ++small, index++);
			} else if (nodeArr[index].value == pivot) {
				index++;
			} else {
				swap(nodeArr, --big, index);
			}
		}
	}

	public static void swap(Node[] nodeArr, int a, int b) {
		Node tmp = nodeArr[a];
		nodeArr[a] = nodeArr[b];
		nodeArr[b] = tmp;
	}

    // 方法2
	public static Node listPartition2(Node head, int pivot) {
		Node sH = null; // small head
		Node sT = null; // small tail
		Node eH = null; // equal head
		Node eT = null; // equal tail
		Node mH = null; // big head
		Node mT = null; // big tail
		Node next = null; // save next node
		// every node distributed to three lists
		while (head != null) {
			next = head.next;
			head.next = null;
			if (head.value < pivot) {
			    // 小於節點為空,當前節點即做頭又做尾
				if (sH == null) {
					sH = head;
					sT = head;
				// 老的尾節點指向當前節點,老的尾變成當前節點
				} else {
					sT.next = head;
					sT = head;
				}
			} else if (head.value == pivot) {
				if (eH == null) {
					eH = head;
					eT = head;
				} else {
					eT.next = head;
					eT = head;
				}
			} else {
				if (mH == null) {
					mH = head;
					mT = head;
				} else {
					mT.next = head;
					mT = head;
				}
			}
			head = next;
		}
		// 小於區域的尾巴,連等於區域的頭,等於區域的尾巴連大於區域的頭
		if (sT != null) { // 如果有小於區域
			sT.next = eH;
			eT = eT == null ? sT : eT; // 下一步,誰去連大於區域的頭,誰就變成eT
		}
		// 上面的if,不管跑了沒有,et
		// all reconnect
		if (eT != null) { // 如果小於區域和等於區域,不是都沒有
			eT.next = mH;
		}
		return sH != null ? sH : (eH != null ? eH : mH);
	}

	public static void printLinkedList(Node node) {
		System.out.print("Linked List: ");
		while (node != null) {
			System.out.print(node.value + " ");
			node = node.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {
		Node head1 = new Node(7);
		head1.next = new Node(9);
		head1.next.next = new Node(1);
		head1.next.next.next = new Node(8);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(2);
		head1.next.next.next.next.next.next = new Node(5);
		printLinkedList(head1);
		// head1 = listPartition1(head1, 4);
		head1 = listPartition2(head1, 5);
		printLinkedList(head1);

	}

}

1.1.4 面試題三

一種特殊的單鏈表結構如下:

public static class Node {
		public int value;
		public Node next;
		public Node rand;

		public Node(int data) {
			this.value = data;
		}
	}

rand指標式單鏈表節點結構中新增加的指標,rand可能指向連結串列中的任意一個節點,也可能為null。給定一個由Node節點型別組成的無環單鏈表節點head。請實現一個函式完成這個連結串列的複製,並返回複製的新連結串列的頭結點。

要求時間複雜度為O(N),額外空間複雜度為O(1)

  1. 雜湊表方法(筆試推薦)

第一步遍歷,把所有節點加入到Map<Node, Node>表示老節點到克隆出來的節點對映

第二步遍歷,查map找到克隆節點,最後返回頭結點

  1. 不用雜湊表的方法,人為構造對應關係(面試推薦)

第一步:每個節點遍歷的時候克隆出來一個新的節點加入到當前節點和其next節點的中間

第二步:此時經過第一步所有節點和其克隆節點都是串在一起的,依次拿出當前節點和其克隆節點,當前節點的rand指標指向的節點的克隆節點給當前節點克隆的節點的rand節點指向的節點。

第三步:此時老節點的rand指標沒變化,克隆節點的rand指標也都指向了對應的克隆節點。此時在大的連結串列上分離出來原連結串列和克隆連結串列

package class06;

import java.util.HashMap;

public class Code04_CopyListWithRandom {

	public static class Node {
		public int value;
		public Node next;
		public Node rand;

		public Node(int data) {
			this.value = data;
		}
	}

    // 方法1
	public static Node copyListWithRand1(Node head) {
		HashMap<Node, Node> map = new HashMap<Node, Node>();
		Node cur = head;
		while (cur != null) {
		    // 當前節點,克隆出來一個相同值的新節點加入字典
			map.put(cur, new Node(cur.value));
			cur = cur.next;
		}
		// 當前節點從頭開始
		cur = head;
		while (cur != null) {
			// cur 老
			// map.get(cur) 新
			map.get(cur).next = map.get(cur.next);
			map.get(cur).rand = map.get(cur.rand);
			cur = cur.next;
		}
		// 返回head對應的克隆節點
		return map.get(head);
	}

    // 方法二
	public static Node copyListWithRand2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		Node next = null;
		// 克隆出來的node放在原本node和next指向的node中間
		// 1 -> 2
		// 1 -> 1' -> 2
		while (cur != null) {
			// cur 老   next 老的下一個
			next = cur.next;
			cur.next = new Node(cur.value);
			cur.next.next = next;
			cur = next;
		}
		cur = head;
		Node curCopy = null;
		// set copy node rand
		// 1 -> 1' -> 2 -> 2'
		// 設定新的克隆節點間的rand節點
		while (cur != null) {
			// cur 老
			// cur.next => 新的 copy出來的節點
			next = cur.next.next;
			curCopy = cur.next;
			curCopy.rand = cur.rand != null ? cur.rand.next : null;
			cur = next;
		}
		// 老的頭結點:head 新克隆出來的頭結點: head.next
		Node res = head.next;
		cur = head;
		// split,分離原本節點組成的連結串列和克隆節點組成的連結串列
		while (cur != null) {
			next = cur.next.next;
			curCopy = cur.next;
			cur.next = next;
			curCopy.next = next != null ? next.next : null;
			cur = next;
		}
		return res;
	}

	public static void printRandLinkedList(Node head) {
		Node cur = head;
		System.out.print("order: ");
		while (cur != null) {
			System.out.print(cur.value + " ");
			cur = cur.next;
		}
		System.out.println();
		cur = head;
		System.out.print("rand:  ");
		while (cur != null) {
			System.out.print(cur.rand == null ? "- " : cur.rand.value + " ");
			cur = cur.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {
		Node head = null;
		Node res1 = null;
		Node res2 = null;
		printRandLinkedList(head);
		res1 = copyListWithRand1(head);
		printRandLinkedList(res1);
		res2 = copyListWithRand2(head);
		printRandLinkedList(res2);
		printRandLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(4);
		head.next.next.next.next = new Node(5);
		head.next.next.next.next.next = new Node(6);

		head.rand = head.next.next.next.next.next; // 1 -> 6
		head.next.rand = head.next.next.next.next.next; // 2 -> 6
		head.next.next.rand = head.next.next.next.next; // 3 -> 5
		head.next.next.next.rand = head.next.next; // 4 -> 3
		head.next.next.next.next.rand = null; // 5 -> null
		head.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4

		printRandLinkedList(head);
		res1 = copyListWithRand1(head);
		printRandLinkedList(res1);
		res2 = copyListWithRand2(head);
		printRandLinkedList(res2);
		printRandLinkedList(head);
		System.out.println("=========================");

	}

}

1.1.5 面試題四

該問題和約舍夫環問題是連結串列問題的比較難的問題

題目描述:給定兩個可能有環也可能無環的單鏈表,頭結點head1和head2。請實現一個函式,如果兩個連結串列相交,請返回相交的第一個節點。如果不相交,返回null

要求:如果兩個連結串列長度之和為N,時間複雜度請達到O(N),額為空間複雜度請達到O(1)

思路:由於是單鏈表,則一旦成環就結束,出不來,因為每個節點只有一個Next指標

  1. 方法一:用set把每個節點的記憶體地址放到set裡面,如果存在相同的記憶體在set中存在,則就是第一個成環的節點

  2. 用快慢指標對連結串列遍歷,那麼快慢指標一定會相遇,能相遇就說明存在環。然後讓慢指標停在相遇的位置,快指標回到頭結點。快指標和慢指標再出發且快指標也變成一次走一步和滿指標相同,再次相遇的節點就是成環節點

package class06;

public class Code05_FindFirstIntersectNode {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getIntersectNode(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		// head1的第一個入環節點
		Node loop1 = getLoopNode(head1);
		// head2的第一個入環節點
		Node loop2 = getLoopNode(head2);
		// 兩個無環連結串列是否相交的情況
		// 由於每個節點只有一個next指標,則如果兩個無環相交則相交之後就只剩下公共部分
		// 方法1把第一條連結串列放到set中,第二個連結串列依次查在不在該set中,第一個找到的就是
		// 方法2
		// 把連結串列1走到尾結點end1,記錄長度l1
		// 把連結串列1走到尾結點end2,記錄長度l2
		// 如果end1和end2的記憶體地址不同一定不相交
		// 如果end1==end2,則(1)長的連結串列從頭結點先走保證和短連結串列相同長度的位置,再以此往下走,第一次相同節點
		// (2)則依次從尾結點出發,找第一次出現記憶體地址不相同的那個節點,該節點的next節點就是第一次相交的節點
		if (loop1 == null && loop2 == null) {
			return noLoop(head1, head2);
		}
		
		// 一個為空,另外一個不為空不可能相交。兩個都不為空的情況下共用一個環
		// 
		if (loop1 != null && loop2 != null) {
			return bothLoop(head1, loop1, head2, loop2);
		}
		return null;
	}

	// 找到連結串列第一個入環節點,如果無環,返回null
	public static Node getLoopNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		// n1 慢  n2 快
		Node n1 = head.next; // n1 -> slow
		Node n2 = head.next.next; // n2 -> fast
		while (n1 != n2) {
			if (n2.next == null || n2.next.next == null) {
				return null;
			}
			n2 = n2.next.next;
			n1 = n1.next;
		}
		// 能相遇則跳出while,快指標回到開頭,滿指標停在原地
		n2 = head; // n2 -> walk again from head
		while (n1 != n2) {
		    // 此時快慢指標每次移動相同步數
			n1 = n1.next;
			n2 = n2.next;
		}
		return n1;
	}

	// 如果兩個連結串列都無環,返回第一個相交節點,如果不想交,返回null
	public static Node noLoop(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		Node cur1 = head1;
		Node cur2 = head2;
		int n = 0;
		while (cur1.next != null) {
			n++;
			cur1 = cur1.next;
		}
		while (cur2.next != null) {
			n--;
			cur2 = cur2.next;
		}
		if (cur1 != cur2) {
			return null;
		}
		// n  :  連結串列1長度減去連結串列2長度的值
		cur1 = n > 0 ? head1 : head2; // 誰長,誰的頭變成cur1
		cur2 = cur1 == head1 ? head2 : head1; // 誰短,誰的頭變成cur2
		n = Math.abs(n);
		while (n != 0) {
			n--;
			cur1 = cur1.next;
		}
		while (cur1 != cur2) {
			cur1 = cur1.next;
			cur2 = cur2.next;
		}
		return cur1;
	}

	// 兩個有環連結串列,返回第一個相交節點,如果不想交返回null
	// head1的入環節點是loop1,head2的入環節點是loop2
	public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
		Node cur1 = null;
		Node cur2 = null;
		// 類似第一種都無環的情況
		if (loop1 == loop2) {
			cur1 = head1;
			cur2 = head2;
			int n = 0;
			while (cur1 != loop1) {
				n++;
				cur1 = cur1.next;
			}
			while (cur2 != loop2) {
				n--;
				cur2 = cur2.next;
			}
			cur1 = n > 0 ? head1 : head2;
			cur2 = cur1 == head1 ? head2 : head1;
			n = Math.abs(n);
			while (n != 0) {
				n--;
				cur1 = cur1.next;
			}
			while (cur1 != cur2) {
				cur1 = cur1.next;
				cur2 = cur2.next;
			}
			return cur1;
		} else {
		    //否則,找第一個成環節點轉回自身的過程中遇到loop2,則相交,否則不相交
			cur1 = loop1.next;
			while (cur1 != loop1) {
				if (cur1 == loop2) {
					return loop1;
				}
				cur1 = cur1.next;
			}
			return null;
		}
	}

	public static void main(String[] args) {
		// 1->2->3->4->5->6->7->null
		Node head1 = new Node(1);
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(7);

		// 0->9->8->6->7->null
		Node head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println(getIntersectNode(head1, head2).value);

		// 1->2->3->4->5->6->7->4...
		head1 = new Node(1);
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(7);
		head1.next.next.next.next.next.next = head1.next.next.next; // 7->4

		// 0->9->8->2...
		head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next; // 8->2
		System.out.println(getIntersectNode(head1, head2).value);

		// 0->9->8->6->4->5->6..
		head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println(getIntersectNode(head1, head2).value);

	}

}

1.1.6 面試題五

題目描述:能不能不給單鏈表的頭結點,只給想要刪除的節點,就能做到在連結串列上把這個點刪掉?

  1. 抖機靈的做法,1->2->3->4->5->null,給定3。那麼根據記憶體地址找到3這個節點,把3下個節點賦值給自身變成4,再把自身的下一個指標指向下下個值5即可。1->2->4->5->null。缺點沒把原始節點刪除,只是改變了值,記憶體地址沒被刪掉而是刪掉了需要刪除節點的下一個記憶體地址。該方法無法刪除連結串列的最後一個節點

實質上不給頭結點,無法刪除給定的節點。沒有頭結點,沒法準確的連指標

    
package class06;

public class Test {

	public static class Node{
		public int value;
		public Node next;
		public Node(int v) {
			value = v;
		}
	}
	
	public static void main(String[] args) {
		Node a = new Node(1);
		Node b = new Node(2);
		Node c = new Node(3);
		
		a.next = b;
		b.next = c;
		// 實質上這裡置為空沒用,只是把Java棧中的變數不指向堆中的3節點
		//堆中的結構沒改變,3節點並沒有被刪除
		c = null;
		
	}
	
}