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[LeetCode]19.Remove Nth Node From End of List
【題目】
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.
【題意】
給你一個鏈表，去除掉倒數第n個節點。
【分析】
設兩個指針 p; q，讓 q 先走 n 步，然後 p 和 q 一起走，直到 q 走到尾節點，刪除 p->next 即可。
【代碼】
/*********************************
* 日期：2014-01-29
* 作者：SJF0115
* 題號: Remove Nth Node From End of List
* 來源：https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
* 結果：AC
* 來源：LeetCode
* 總結：
**********************************/
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
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 || n <= 0){
return head;
}
ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
dummy->next = head;
ListNode *temp = NULL;
ListNode *pre = dummy,*cur = dummy;
//cur先走N步
for(int i = 0;i < n;i++){
cur = cur->next;
}
//同步前進直到cur到最後一個節點
while(cur->next != NULL){
pre = pre->next;
cur = cur->next;
}
//刪除pre->next
temp = pre->next;
pre->next = temp->next;
delete temp;
return dummy->next;
}
};
int main() {
Solution solution;
int A[] = {1,2,3,4,5};
ListNode *head = (ListNode*)malloc(sizeof(ListNode));
head->next = NULL;
ListNode *node;
ListNode *pre = head;
for(int i = 0;i < 5;i++){
node = (ListNode*)malloc(sizeof(ListNode));
node->val = A[i];
node->next = NULL;
pre->next = node;
pre = node;
}
head = solution.removeNthFromEnd(head->next,4);
while(head != NULL){
printf("%d ",head->val);
head = head->next;
}
return 0;
}
最後更新：2017-04-03 12:54:51