[LeetCode]24.Swap Nodes in Pairs
2014-01-29 17:02
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【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.
【题意】
给定一个链表,交换每两个相邻的节点,并返回它的头。你的算法应该使用常数空间。您不得修改在列表中的值,只有节点本身是可以改变的。
【分析】
无【代码】
/********************************* * 日期:2014-01-29 * 作者:SJF0115 * 题号: 24.Swap Nodes in Pairs * 来源:http://oj.leetcode.com/problems/swap-nodes-in-pairs/ * 结果:AC * 来源:LeetCode * 总结: **********************************/ #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个节点采用交换 while(cur != NULL && cur->next != NULL){ nodeA = cur; nodeB = cur->next; //交换 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; }
【代码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; //统计节点个数 int count = 0; while(pre->next != NULL){ pre = pre->next; count++; } //反转次数 int rCount = count / k; int index = 0; //反转元素的前一个 pre = dummy; //反转元素第一个即翻转后的尾元素 tail = dummy->next; //共进行rCount次反转 while(index < rCount){ int i = k - 1; //反转 while(i > 0){ //删除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; } };
利用:LeetCode之Reverse Nodes in k-Group
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