leetcode -day30 Reverse Linked List II

1、



Reverse Linked List II 

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m =
2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

分析：開始看题目以为是仅仅交换两个指定位置的值。后来写出代码来出错。才发现是翻转位置m和n直接的链表，我的基本思路是先用双指针法找到要翻转链表的位置，将链表分成三段，要翻转的前一段，中间呀翻转的段，和剩下的段。最后写出代码来特别繁琐。例如以下所看到的：

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(m < 1 || m >= n){
            return head;
        }
        //双指针法找到要翻转的链表段
        ListNode* node1 = head;
        ListNode* node2 = head;
        int dis = n - m;
        int i = 0;
        for(; i<dis && node2; ++i){
            node2 = node2->next;
        }
        if(i<dis){
            return head;
        }
        while(i<n-2 && node2){
            node1 = node1->next;
            node2 = node2->next;
            ++i;
        }
        //node3为要翻转的链表段的開始结点
        ListNode* node3 = node1->next;
        if(m == 1){
            node1 = NULL;
            node3 = head;
        }else{
            node1->next = NULL;
            node2 = node2->next;
            if(!node2){
                return head;
            }
            ++i;
        }
        //node4为剩下的链表
        ListNode* node4 = NULL;
        node4 = node2->next;
        node2->next = NULL;
        //假设链表长度大于n时。能够进行
        if(i == n-1){
            ListNode* newHead = reverseList(node3); //翻转中间链表
            //连接三段链表
            node3->next = node4;
            if(m !=1 ){
                node1->next = newHead;
                return head;
            }else{
                return newHead;
            }
        }
        return head;
    }
    ListNode* reverseList(ListNode* head){
        ListNode* node1 = NULL;
        ListNode* node2 = head;
        ListNode* tempNode = NULL;
        while(node2){
            tempNode = node2->next;
            node2->next = node1;
            node1 = node2;
            node2 = tempNode;
        }
        return node1;
    }
};

改进：上述代码非常繁琐。搜了别人的代码，非常简洁，问题在于上述对中间链表进行了两次遍历，缩短为一次遍历。可缩减代码。上列代码没有考虑异常情况，比方链表长度<m或者<n等情况。

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;

        ListNode *q = NULL;
        ListNode *p = head;
        for(int i = 0; i < m - 1; i++)
        {
            q = p;
            p = p->next;
        }

        ListNode *end = p;
        ListNode *pPre = p;
        p = p->next;
        for(int i = m + 1; i <= n; i++)
        {
            ListNode *pNext = p->next;

            p->next = pPre;
            pPre = p;
            p = pNext;
        }

        end->next = p;
        if (q)
            q->next = pPre;
        else
            head = pPre;

        return head;
    }
};