lintcode-36-翻转链表 II

36-翻转链表 II

翻转链表中第m个节点到第n个节点的部分

注意事项

m，n满足1 ≤ m ≤ n ≤ 链表长度

样例

给出链表1->2->3->4->5->null， m = 2 和n = 4，返回1->4->3->2->5->null

挑战

在原地一次翻转完成

标签

链表

思路

借助2个空的节点，置于链表头部，一次遍历链表，找到翻转子链尾部subEnd，不翻转子链尾部lastEnd，头部nextHead，以及当前节点nowNode，在遍历时，将nowNode插入lastEnd与subEnd之间。最后，将翻转子链与nextHead相连。

code

/**
 * Definition of singly-linked-list:
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The head of linked list.
     * @param m: The start position need to reverse.
     * @param n: The end position need to reverse.
     * @return: The new head of partial reversed linked list.
     */
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // write your code here
        
        ListNode newHead1(-1), newHead2(-1);
        newHead1.next = &newHead2;
        newHead2.next = head;

        ListNode *nowNode=head, *subHead=head, *subEnd=NULL, *lastEnd=&newHead1, *nextHead=NULL;
        int i=1;
        while(nowNode != NULL) {
            if(i <= m) {
                lastEnd = lastEnd->next;
                subEnd = nowNode;
                nextHead = nowNode;
                nowNode = nowNode->next;
                i++;
            }
            else if(i <= n) {
                ListNode *temp = nowNode->next;
                nextHead->next = NULL;
                lastEnd->next = nowNode;
                lastEnd->next->next = subEnd;
                subEnd = nowNode;
                nowNode = temp;
                i++;
            }
            else {
                nextHead->next = nowNode;
                break;
            }
        }

        if(m == 1) {
            return lastEnd->next;
        }
        else {
            return head;
        }
    }
};