leetcode 76: Reverse Linked List II

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本文介绍了一种算法，用于在给定范围内反转链表的一部分。该方法可在原地完成且只需一次遍历。提供了C++及Java实现代码。

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Reverse Linked List IIJun 27 '12

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:

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        //first let's try reverse linkedlist.
        
        //12345  2,4
        
        ListNode ** pp = &head;
        ListNode * p = head;
        ListNode * pre = NULL;
        ListNode * nxt = NULL;
        n = n-m+1;
        
        while(p!=NULL && --m>0) {
            pp = &(p->next);
            p = p->next;
        }
        
        if(m!=0) return head;
        
        while( p!=NULL && n-->0) {
            nxt = p->next;
            p->next = pre;
            pre = p;
            p = nxt;
        }
        
        ListNode * t = *pp;
        *pp = pre;
        t->next = nxt;
        
        return head;
    }
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if( head==null | m==n) return head;
        
        ListNode start = new ListNode(0);
        start.next = head;
        head = start;
        n=n-m+1; 
        
        while(--m>0 && start!=null) start = start.next;
        if(start==null) return head.next;
        
        ListNode cur = start.next;
        ListNode pre = null;
        ListNode nxt = null;
        
        while(cur!=null && n-->0) {
            nxt = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nxt;
        }
        start.next.next = cur;
        ListNode t = start.next;
        start.next = pre;
        start = t;
        
        return head.next;
    }
}