Swap Nodes in Pairs-优快云博客

本文提供两种方法解决链表中每两个相邻节点交换的问题：递归和迭代。通过这两种方法，可以在常数空间内完成节点交换而不改变链表中的值。

题目

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

思路一：递归

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head==NULL || head->next==NULL)
            return head; 
        ListNode * pre = head;
        ListNode * cur = head->next;
        if(pre && cur)
        {
            pre->next = swapPairs(cur->next);
            cur->next = pre;
        }
        return cur;
    }
    
};

类似于 Remove Duplicates from Sorted List II

最新 java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode pre = head;
        head = head.next;
        
        pre.next = head.next;
        head.next = pre;
        
        pre.next = swapPairs(pre.next);
        return head;
    }
}

思路二：迭代

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
    if (!head || !head->next) 
        return head;
    ListNode *ppre = NULL, *prev = head, *curr = head->next;

    while (curr) { //left at least 2 node
        prev->next = curr->next;
        curr->next = prev;
        if (ppre) 
            ppre->next = curr;
        else 
            head = curr;

        ppre = prev;
        prev = prev->next;
        if (prev) 
            curr = prev->next; //left at least 1 node
        else 
            curr = NULL;            //none left
    }
    return head;
}
    
};