力扣 题目92- 反转链表 II

题目

题解

参考 力扣 题目25-- K 个一组翻转链表 的做法使用栈即可

代码

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
struct ListNode {
    int val;
    ListNode* next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode* next) : val(x), next(next) {}
};

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        if (head == nullptr || left == right) {
            return head;
        }
        //创造一个无关的结点并连接原来的链表 就可以使原本的头结点 普通化
        ListNode* result = new(ListNode);
        result->next = head;
        //记录反转开头位置(即left-1)
        ListNode* begin= head;
        stack<ListNode*> reversal;
        //重置头结点为新
        head = result;
        for (int i = 0; i<= right; i++) {
            //记录栈
            if (left <= i&& i<= right) {
                reversal.push(head);
            }
            //记录反转开头位置(即left-1)
            if (i == left - 1)
            {
                begin = head;
            }
            head = head->next;
        }
        //栈出
        while (!reversal.empty())
        {
            begin->next = reversal.top();
            begin = begin->next;
            reversal.pop();
        }
        //接上反转后链表
        begin->next = head;
        return result->next;
    }
};


void listnum(ListNode& list, vector<int>& vectorlist) {
    ListNode* headlist = &list;
    for (int i = 0; i < vectorlist.size(); i++) {
        ListNode* p;
        p = (struct ListNode*)malloc(sizeof(struct ListNode*));
        p->val = vectorlist[i];
        headlist->next = p;
        headlist = headlist->next;
    }
    headlist->next = NULL;
}
int main() {
    Solution sol;
    ListNode list1(3);
    vector<int> vectorlist1 = { 5 };
    listnum(list1, vectorlist1);
    ListNode* list = sol.reverseBetween(&list1,1,2);
    for (int i = 0; list != NULL; i++) {
        cout << list->val << endl;
        list = list->next;
    }
}