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[LeetCode]24.Swap Nodes in Pairs

【題目】

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.

【題意】

給定一個鏈表，交換每兩個相鄰的節點，並返回它的頭。

你的算法應該使用常數空間。您不得修改在列表中的值，隻有節點本身是可以改變的。

【分析】

無

【代碼】

/*********************************
*   日期：2014-01-29
*   作者：SJF0115
*   題號: 24.Swap Nodes in Pairs
*   來源：https://oj.leetcode.com/problems/swap-nodes-in-pairs/
*   結果：AC
*   來源：LeetCode
*   總結：
**********************************/
#include <iostream>
#include <stdio.h>
#include <algorithm>
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){
            return head;
        }

        ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
        dummy->next = head;

        ListNode *nodeA,*nodeB = NULL;

        ListNode *pre = dummy,*cur = head;
        //至少2個節點采用交換
        while(cur != NULL && cur->next != NULL){
            nodeA = cur;
            nodeB = cur->next;
            //交換
            nodeA->next = nodeB->next;
            nodeB->next = nodeA;
            pre->next = nodeB;
            //更新pre,cur
            pre = nodeA;
            cur = nodeA->next;
        }
        return dummy->next;
    }
};
int main() {
    Solution solution;
    int A[] = {1,2,3,4,5};
    ListNode *head = (ListNode*)malloc(sizeof(ListNode));
    head->next = NULL;
    ListNode *node;
    ListNode *pre = head;
    for(int i = 0;i < 4;i++){
        node = (ListNode*)malloc(sizeof(ListNode));
        node->val = A[i];
        node->next = NULL;
        pre->next = node;
        pre = node;
    }
    head = solution.swapPairs(head->next);
    while(head != NULL){
        printf("%d ",head->val);
        head = head->next;
    }
    return 0;
}

【代碼2】

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        return reverseKGroup(head,2);
    }
private:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if (head == NULL || k < 2){
            return head;
        }

        ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
        dummy->next = head;

        ListNode *pre = dummy,*cur = NULL,*tail = NULL;
        //統計節點個數
        int count = 0;
        while(pre->next != NULL){
            pre = pre->next;
            count++;
        }
        //反轉次數
        int rCount = count / k;
        int index = 0;
        //反轉元素的前一個
        pre = dummy;
        //反轉元素第一個即翻轉後的尾元素
        tail = dummy->next;
        //共進行rCount次反轉
        while(index < rCount){
            int i = k - 1;
            //反轉
            while(i > 0){
                //刪除cur元素
                cur = tail->next;
                tail->next = cur->next;
                //插入cur元素
                cur->next = pre->next;
                pre->next = cur;
                i--;
            }
            pre = tail;
            tail = pre->next;
            index++;
        }
        return dummy->next;
    }
};

最後更新：2017-04-03 12:54:51

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