力扣算法题—092反转链表2

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

#include "_000库函数.h"


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


//只能扫描一遍
//将中间要反转的数取出来在放入链表中
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (!head || m <= 0 || n <= 0 || n <= m)return head;
        //加头
        ListNode*p = new ListNode(-1);
        p->next = head;
        head = p;

        stack<int>s;
        ListNode*pre = new ListNode(0);
        while (p && n >0) {
            --m;
            if (m == 0)pre = p;            
            p = p->next;
            if (p && m <= 0)
                s.push(p->val);
            --n;
        }
        if (p == NULL)return head->next;//m,n超过了链表长度
        pre->next = p->next;
        while (!s.empty()) {
            ListNode*q = new ListNode(0);
            q->val = s.top();
            q->next = pre->next;            
            pre->next = q;
            pre = q;
            s.pop();
        }
        return head->next;
    }
};

//走一步反转一个数据
//不用对m,n的大小进行判断
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        ListNode *dummy = new ListNode(-1), *pre = dummy;
        dummy->next = head;
        for (int i = 0; i < m - 1; ++i) pre = pre->next;
        ListNode *cur = pre->next;
        for (int i = m; i < n; ++i) {//用交换法
            ListNode *t = cur->next;
            cur->next = t->next;
            t->next = pre->next;
            pre->next = t;
        }
        return dummy->next;
    }
};


void T092() {
    Solution s;
    vector<int>v;
    ListNode *head = new ListNode(0);
    ListNode *p = head;
    v = { 1,2,3,4,5 };
    for (auto a : v) {
        ListNode *q = new ListNode(0);
        q->val = a;
        p->next = q;
        p = q;
    }
    p = head->next;
    while (p) {
        cout << p->val << "->";
        p = p->next;
    }
    cout << endl;


    p = s.reverseBetween(head->next, 2, 6);
    while (p) {
        cout << p->val << "->";
        p = p->next;
    }
    cout << endl;
}