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  • 【leetcode】Swap Nodes in Pairs (middle)

    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.

    //start time = 9:54
    //end time = 10:16
    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <queue>
    #include <stack>
    using namespace std;
    
     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;
            //头指针转换
            ListNode * newhead = head->next;
            head->next = newhead->next;
            newhead->next = head;
    
            ListNode * pre = newhead->next; //之前转换完的最后一个
            ListNode * cur = NULL; //一对中的前一个
            ListNode * next = NULL;//一对中的后一个
            while(pre->next != NULL && pre->next->next != NULL)
            {
                cur = pre->next;
                next = cur->next;
                pre->next = next;
                cur->next = next->next;
                next->next = cur;
    
                pre = cur;
            }
    
            return newhead;
        }
    };

    发现多出了c的选项 用c写的时候 ListNode前面要加 struct 修饰 而C++不用 注意区别 时间上C需要1ms 而C++需要6ms

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     struct ListNode *next;
     * };
     */
    struct ListNode *swapPairs(struct ListNode *head) {
        if(head == NULL || head->next == NULL) return head;
            //头指针转换
            struct ListNode * newhead = head->next;
            head->next = newhead->next;
            newhead->next = head;
    
            struct ListNode * pre = newhead->next; //之前转换完的最后一个
            struct ListNode * cur = NULL; //一对中的前一个
            struct ListNode * next = NULL;//一对中的后一个
            while(pre->next != NULL && pre->next->next != NULL)
            {
                cur = pre->next;
                next = cur->next;
                pre->next = next;
                cur->next = next->next;
                next->next = cur;
    
                pre = cur;
            }
    
            return newhead;
    }
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  • 原文地址:https://www.cnblogs.com/dplearning/p/4310320.html
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