[LeetCode]78. Remove Nth Node From end of List删除链表中倒数第N个节点

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.

 

Subscribe to see which companies asked this question

 

解法1：首先扫描一遍链表，统计共有多少个元素；然后从头开始遍历链表，直到到达要删除节点的前一个节点。因为题目限定了输入n总是有效的，所以不需要考虑n大于链表长度或者n小于等于0这些情况。

/**
 * 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 == NULL) return NULL;
        ListNode* p = head;
        int num = 0;
        while(p != NULL) {
            ++num;
            p = p->next;
        }
        if(num - n == 0) { // 删除的是头节点
            ListNode* del = head;
            head = head->next;
            delete del;
        }
        else {
            p = head;
            for(int i = 0; i < num - n - 1; ++i)
                p = p->next;
            ListNode* del = p->next;
            p->next = p->next->next;
            delete del;
        }
        return head;
    }
};

 

解法2：一趟遍历解决问题。因为删除某个节点需要改动其前一个节点的next指针，因此我们先找到要删除节点的前一个节点。设置快慢两个指针，初始时都指向头节点。在快指针向前移动N步之后（注意，若此时快指针到达了链表的尾部，即为NULL，则说明删除的是头节点，需要单独考虑），两个指针同时向后移动，直到快指针是尾节点。

/**
 * 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) {
        if(head == NULL) return NULL;
        ListNode* p1 = head;
        ListNode* p2 = head;
        for(int i = 0; i < n; ++i)
            p1 = p1->next;
        if(p1 == NULL) { //删除的是头节点
            ListNode* del = head;
            head = head->next;
            delete del;
            return head;
        }
        while(p1->next != NULL) {
            p1 = p1->next;
            p2 = p2->next;
        }
        ListNode* del = p2->next;
        p2->next = p2->next->next;
        delete del;
        return head;
    }
};