[leetcode 27] Remove Element

题目：

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.

思路：

1.用result（初始为0）做下标，遍历数组，如果当前值等于给定值，则跳过，否则A[result++]=A[i];

代码：

class Solution{
public:
	int removeElement(int A[],int n,int elem)
	{
		int result=0;
		for(int i=0;i!=n;++i)
		{
		    if(A[i]!=elem)
			{
			    A[result++]=A[i];
			}
		}
		return result;
	}
};