【leetcode】【单链表】【19】Remove Nth Node From End of List

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本文介绍了一个链表操作的C++实现：删除链表中倒数第N个节点的方法。通过两次遍历的方式，首先统计链表长度，然后定位到待删除节点的前一个节点，进行节点删除操作。提供了创建链表、打印链表和删除节点的具体实现。

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#include<iostream>
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) {
		int numOfNode = 0;
		ListNode* cur = head;
		while (cur){ //统计节点个数
			++numOfNode;
			cur = cur->next;
		}
		int beforeDelete = numOfNode - n;
		ListNode* temp = NULL;
		if (beforeDelete == 0){ //删除头结点情况
			temp = head;
			head = head->next;
		}else{ //删除非头结点情况
			cur = head;
			while (--beforeDelete > 0)
				cur = cur->next;
			temp = cur->next;
			cur->next = temp->next;
		}
		delete temp;
		return head;
	}
	ListNode* createList(ListNode* head){
		int numOfNode;
		int value;
		cout << "please input number of listNode:";
		cin >> numOfNode;
		cin >> value;
		head = new ListNode(value);
		ListNode* cur = head;
		for (int i = 1; i < numOfNode; ++i){
			cin >> value;
			ListNode* temp = new ListNode(value);
			cur->next = temp;
			cur = temp;
		}
		return head;
	}
	void printNode(ListNode* head){
		ListNode* cur = head;
		while (cur){
			cout << cur->val << " ";
			cur = cur->next;
		}
		cout << endl;
	}
};

int main(){
	ListNode* head = NULL;
	Solution lst;
	head = lst.createList(head);
	lst.printNode(head);

	head = lst.removeNthFromEnd(head, 2);
	lst.printNode(head);

	system("pause");
	return 0;
}