題目連結

328. 奇偶連結串列

題目描述

解題思路

暴力法

先統計連結串列長度，然後根據連結串列長度決定兩兩交換連結串列中的元素的次數即可。（如果不明白自己畫個圖即可）

拆分為奇偶連結串列在拼接

維護兩個指標 odd 和 even 分別指向奇數節點和偶數節點，初始時 odd = head，even = evenHead。通過迭代的方式將奇數節點和偶數節點分離成兩個連結串列，每一步首先更新奇數節點，然後更新偶數節點。最後把偶數節點拼接在奇數節點後面即可。

AC程式碼

解法一

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null || head.next.next ==null) return head;
        ListNode countNode = head;
        int count = 0;
        while(countNode != null){
            count++;
            countNode = countNode.next;
        }
        //連結串列的經典解法，3指標！！！！！！
        ListNode dum = head;
        ListNode pre = head;
        ListNode cur = head.next;
        ListNode temp = cur.next;
        int tot = 0;
        //根據連結串列長度決定兩兩元素交換的次數
        if(count % 2 == 0) tot = count / 2 - 1;
        else tot = count / 2;
        while(tot > 0){
            int num = tot;
            while(num > 0){
                pre.next = temp;
                cur.next = temp.next;
                temp.next = cur;
                pre = cur;
                if(cur.next != null) cur = cur.next;
                if(cur.next != null) temp = cur.next;
                num--;
            }
            tot--;
            dum = dum.next;
            pre = dum;
            cur = pre.next;
            temp = cur.next;
        }
        return head;
    }
}
 

解法二

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null || head.next.next ==null) return head;
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        while(true){
            odd.next = even.next;
            if(even.next != null) odd = even.next;
            else break;
            if(odd.next != null) even.next = odd.next;
            else break;
            if(odd.next != null) even = odd.next;
        }
        odd.next = evenHead;
        even.next = null;
        return head;
    }
}