[LeetCode]Reverse Linked List II

题目：Reverse Linked List II

逆置链表m到n的之间的节点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL ||m == n)return head;
        int count = 0;
        ListNode *p = head;
        ListNode *pre = NULL;
        while (count < m - 1){//找到m的节点的位置
            pre = p;
            p = p->next;
            count++;
        }
        ListNode *start = p;
        while (count < n - 1){//找到n的节点的位置
            p = p->next;
            count++;
        }
        reverseRecursion(start,p);
        if (pre == NULL){
            return p;
        }
        pre->next = p;
        return head;
    }
private:
    void reverseRecursion(ListNode *p,ListNode *end){//递归的方法逆置链表p
        ListNode *q;
        if (p->next == end){//倒数第二个节点
            q = end->next;
            p->next->next = p;
            p->next = q;
            return;
        }
        reverseRecursion(p->next, end);
        q = p->next->next;//逆置当前节点
        p->next->next = p;
        p->next = q;
    }
};

能用递归就能用栈来实现，

用栈来逆置链表，做法如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL ||m == n)return head;
        int count = 0;
        ListNode *p = head;
        ListNode *pre = NULL;
        while (count < m - 1){
            pre = p;
            p = p->next;
            count++;
        }
        ListNode *start = p;
        while (count < n - 1){
            p = p->next;
            count++;
        }
        if (pre == NULL){
            return reverseNonRecursion(pre, start, p);
        }else{
            reverseNonRecursion(pre,start,p);
            return head;
        }
    }
private:
    ListNode *reverseNonRecursion(ListNode *pre, ListNode *p, ListNode *end){
        stack<ListNode *>stackLN;
        ListNode *q = p;
        while (q != end){//入栈
            stackLN.push(q);
            q = q->next;
        }
        end = end->next;
        if(pre == NULL)pre = q;//得到前一个节点
        else pre->next = q;
        while (!stackLN.empty()){
            p = stackLN.top();
            stackLN.pop();
            q->next = p;//逆置
            q = q->next;
        }
        q->next = end;
        return pre;
    }
};

直接用循环来逆置链表：

1->2->3->4->5;

(m,n)=(2,4);

p=2;pre=1;end=4;

q=5;

1)temp=p;

2)temp->next=q;//2->5

3)p=p->next;//p=3;

4)q=temp;//q=2;

循环1到4步

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL ||m == n)return head;
        int count = 0;
        ListNode *p = head;
        ListNode *pre = NULL;
        while (count < m - 1){
            pre = p;
            p = p->next;
            count++;
        }
        ListNode *start = p;
        while (count < n - 1){
            p = p->next;
            count++;
        }
        if (pre == NULL){
            return reverseNonRecursion2(pre, start, p);
        }else{
            reverseNonRecursion2(pre,start,p);
            return head;
        }
    }
private:
    ListNode *reverseNonRecursion2(ListNode *pre, ListNode *p, ListNode *end){
        ListNode *q = end->next;
        while (p != end){
            ListNode *temp = p;
            p = p->next;
            temp->next = q;
            q = temp;
        }
        p->next = q;
        if (pre == NULL)return p;
        pre->next = p;
        return pre;
    }
};