LeetCode算法系列：24. Swap Nodes in Pairs

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题目描述：

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

Example:

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

Note:

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

算法实现：

这个问题比较简单，只需要定义三个指针扫描链表即可，其中两个指向当前要交换的指针，另一个要指向这两个指针的前一个指针，因为要将其重新链接到前一个指针上

/**
 * 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) {
        if(head == NULL || head -> next == NULL)return head;
        ListNode* f = head, *b = head -> next, *p;
        f -> next = b -> next;
        b -> next = f;
        head = b;
        
        while(f -> next != NULL && f -> next -> next != NULL){
            p = f;
            f = f -> next;
            b = f -> next;
            
            p -> next = b;
            f -> next = b -> next;
            b -> next = f;
        }
        return head;
    }
};