Reverse Linked List II --一道很锻炼链表和指针的题

最新推荐文章于 2024-09-01 21:01:24 发布

Edward_Wong

最新推荐文章于 2024-09-01 21:01:24 发布

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分类专栏： LeetCode 文章标签：链表指针 Leetcode

本文链接：https://blog.youkuaiyun.com/Edward_Wong/article/details/39577445

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本文详细介绍了如何使用C++实现LeetCode中关于反转链表特定区间的算法，包括关键步骤和代码实现。

题目链接：https://oj.leetcode.com/problems/reverse-linked-list-ii/

#include <iostream>
using namespace std;

struct ListNode {
	int val;
	ListNode *next;
	ListNode(): val(0), next(NULL) {}
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	ListNode *reverseBetween(ListNode *head, int m, int n) {
		if(head == NULL || m == n) return head;
		int i;
		ListNode *first = NULL;
		ListNode *temp  = head;
		for(i = 0; i < m - 1; i++)
		{
			first = temp;
			temp  = temp -> next;
		}
		ListNode *end      = temp;
		ListNode *temp_pre = temp;
		temp               = temp -> next;
		for(i = m + 1; i <= n; i++)                  //<span style="font-family: Arial, Helvetica, sans-serif;">主要步骤：将该节点的next指针指向前一个节点，然后temp_pre和temp都往前移</span>
		{
			ListNode *temp_next = temp -> next;
			temp -> next        = temp_pre;
			temp_pre            = temp;
			temp                = temp_next;
		}
		end -> next = temp;                         <span style="font-family: Arial, Helvetica, sans-serif;">//将原来的第m个节点与第n+1个节点连接</span>
		if(first) {                                 <span style="font-family: Arial, Helvetica, sans-serif;">//判断是否从一开始就反向</span>
			first -> next = temp_pre;
		}
		else {
			head = temp_pre;
		}
		return head;
	}
};

int main() {
	int len,m,n,t,i,j;
	cin>>len;
	ListNode *node = new ListNode[len+1], *head;
	head = node;
	for(i = 0; i < len;i++) {
		cin>>t;
		j = i;
		node[i].val  = t;
		node[i].next = &node[++j];
	}
	cin>>m>>n;
	Solution a;
	ListNode *res = a.reverseBetween(head,m,n);
	for(i = 0; i < len; i++) {
		if(i != len - 1) {
			cout<<res -> val<<"->";
			res = res -> next;
		}
		else {
			cout<<res -> val<<endl;
		}
	}
	delete [] node;
	system("PAUSE");
}