LeetCode之“链表”：Remove Nth Node From End of List

　　题目链接

　　题目要求：

　　Given a linked list, remove the nth node from the end of list and return its head.

　　For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

　　Note:

　　Given n will always be valid.
　　Try to do this in one pass.

　　这道题比较常规的想法是先算出链表长度len，再将头指针移动len-n步就可以到达待删节点了。具体程序如下（4ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head)
            return head;
        
        int len = 0;
        ListNode *start = head;
        while(start)
        {
            len++;
            start = start->next;
        }
        
        if(len == n)
        {
            start = head;
            head = head->next;
            delete start;
            start = nullptr;
            return head;
        }
        
        ListNode *preNode = head;
        start = preNode->next;
        int k = 1;
        while(k < len - n)
        {
            preNode = start;
            start = start->next;
            k++;
        }
        
        preNode->next = start->next;
        delete start;
        start = nullptr;
        
        return head;
    }
};

　　另一个比较巧妙的方法就是先定义一个慢指针（初始值为头指针父指针），让头指针移动n步，再让头指针和一个慢指针同时移动直至头指针为空，这时慢指针指向的下一个节点就是待删节点。具体程序日如下（4ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head)
            return head;
        
        ListNode *dummy = new(nothrow) ListNode(INT_MIN);
        assert(dummy);
        dummy->next = head;
        
        for(int i = 0; i < n; i++)
            head = head->next;
        
        ListNode *preNode = dummy;
        while(head)
        {
            head = head->next;
            preNode = preNode->next;
        }
        
        ListNode *delNode = preNode->next;
        preNode->next = delNode->next;
        delete delNode;
        delNode = nullptr;
        
        head = dummy->next;
        delete dummy;
        dummy = nullptr;
        
        return head;
    }
};