LeetCode——Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

» Solve this problem

此题的难点在于只能遍历一次链表。

链表本身包含的信息并不多，只有next和value。

因此想要只遍历一次链表就找到倒数第n个节点只能从指针入手。

受启发于《判断链表是否存在环》一题，我们设置两个指针：P1和P2，P2是P1之后的第n-1个点。

当P2到达尾部时，P1所指即为倒数第n个节点。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL) {
            return NULL;
        }
        ListNode *p1 = head;
        ListNode *p2 = head;
        for (int i = 0; i < n - 1; i++) {
            p2 = p2->next;
        }
        ListNode *prev = NULL;
        while (p2->next != NULL) {
            prev = p1;
            p1 = p1->next;
            p2 = p2->next;
        }
        if (p1 == head) {
            head = p1->next;
            delete p1;
        }
        else {
            prev->next = p1->next;
            delete p1;
        }
        return head;
    }
};