标签：c++   leetcode   array   

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Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.

//vs2012测试
#include<iostream>
#include<algorithm>

using namespace std;

#define n 2

int main()
{
	int A[n];
	int elem;
	for(int i=0; i<n; i++)
	{
		cin>>A[i];
	}
	cin>>elem;

	int start = 0;
	for(int i=0; i<n; i++)
	{
		if(A[i] != elem)
		{
			A[start] = A[i];
			start++;
		}
	}
	//for(int j=start; j<n; j++)
	//	A[j]=0;

	for(int i=0; i<start; i++)
		cout<<A[i];
	cout<<endl;

	cout<<start<<endl;

	return start;
}

//方法一：自测Accepted
class Solution {
    
public:
    int removeElement(int A[], int n, int elem) {
	int length = 0;
	for(int i=0; i<n; i++)
	{
		if(A[i] != elem)
		{
			A[length] = A[i];
			length++;
		}
	}
	for(int j=length; j<n; j++)
		A[j]=0;



	//for(int i=0; i<n; i++)
	//	cout<<A[i];
	//cout<<endl;

	//cout<<length<<endl;

	return length;
    }
};

leetcode_27_Remove Element

标签：c++   leetcode   array   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/43764445