常用演算法的總結——連結串列
阿新 • • 發佈:2021-01-01
連結串列常用演算法:快慢指標、利用HashMap、棧
如:
1 快慢指標
1) 輸入連結串列頭節點,奇數長度返回中點,偶數長度返回上中點
2) 輸入連結串列頭節點,奇數長度返回中點,偶數長度返回下中點
3) 輸入連結串列頭節點,奇數長度返回中點前一個,偶數長度返回上中點前一個
4) 輸入連結串列頭節點,奇數長度返回中點前一個,偶數長度返回下中點前一個
第一題程式碼如下,其他同理:
// head頭 public static Node mi d0rUpMi dNode(Node head) { if (head == nu11|I head.next == nu11Il| head.next.next == nu1l) {return head;
} Node slow = head . next; Node fast = head. next . next; while (fast.next != nu11 && fast. next.next != nu11) { slow = slow. next; fast = fast. next. next;
} return slow;
}
2給定一個單鏈表的頭節點head,請判斷該連結串列是否為迴文結構。
1) 雜湊表方法特別簡單——利用棧
public static boolean isPalindrome1 (Node head) { Stack<Node> stack = new Stack<Node>( ); Node cur = head; while (cur != nu1l) { stack. push(cur); cur = cur. next; } while (head != nu1l) { if (head.value != stack. pop().value) { return false; } head = head. next; } return true; }
2) 改原連結串列的方法就需要注意邊界了
3) 利用快慢指標的演算法:先讓快指標到終點,慢指標到中點,然後讓指向中點的慢指標指向null,後半部分全部倒轉,後面的指向前面的,最後由終點和起點開始往中間遍歷,只要有一步不相等就返回false,最後由如果需要恢復則將其恢復成原樣,程式碼如下(包含了最後恢復成原樣的步驟):
public static boolean isPalindrome3(Node head) { if (head == null | head.next == null) { return true; } Node n1 = head; 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 } n2 = n1.next; // n2 -> right part first node n1.next = null; // mid.next -> null Node n3 = null; while (n2 != null) { // right part convert n3 = n2.next; // n3 -> save next node n2.next = n1; // next of right node convert n1 = n2;//n1move n2 = n3;//n2move } 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; while (n1 != null) { // recover list n2 = n1. next; n1.next = n3; n3 = n1; n1 = n2; } return res; }