[LeetCode] 92. Reverse Linked List II

92. 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

这个题目和直接翻转链表类似，麻烦的地方就是定位出需要翻转链表的头和尾，要注意的条件特别多。请看源码

源码

/** * 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) return NULL;        if(head->next == NULL) return head;        if(m == n || n < 2) return head; // m <= n m=n时不需要翻转，n<2时即m=n=1也不需要翻转        //定位需要翻转链表的头尾指针以及头指针的前一个指针，尾指针的后一个指针        ListNode *pmPre = NULL, *pm = NULL, *pn = NULL,*pnNext = NULL;        ListNode *pList = head;        int i = 1;        while(i <= (n+1)) {            if(i == (m-1))pmPre = pList; // m = 1时，说明链表第一个元素开始，那头指针前面元素就是空指针了            if(i == m)pm = pList;            if(i == n)pn = pList;            if(i == n+1)pnNext = pList; // 尾指针后一个指针            i++;            if(pList)pList = pList->next;        }        // 为了翻转指针，先将尾指针的next置空        pn->next = NULL;        reverseList(pm); // 递归翻转指针        if(pmPre) // 如果pmPre不为空，将pmPre的下一个元素置成翻转后的第一个元素            pmPre->next = pn;        pm->next = pnNext; // 将翻转后的尾指着的next置成pnNext        return m == 1 ? pn : head;    }    ListNode* reverseList(ListNode* head) {        if(!head)return NULL;        if(head->next == NULL)return head;        ListNode* pNode = head->next;        ListNode* pHead = reverseList(pNode);        pNode->next = head;        head->next = NULL;        return pHead;    }};