Leetcode: 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.
这道题比较简单直接看代码，这里多写几个版本的作参考！ C++版本
class Solution
{
public:
int removeElement(int A[], int n, int elem)
{
int pt = 0;
for (int i = 0; i < n; i++)
{
if (A[i] != elem) A[pt++] = A[i];
}
return pt;
}
};C#版本：
public class Solution
{
public int RemoveElement(int[] A, int elem)
{
int pt = 0;
for (int i = 0; i < A.Length; i++)
{
if (A[i] != elem) A[pt++] = A[i];
}
return pt;
}
}Python版本：
class Solution:
# @param A a list of integers
# @param elem an integer, value need to be removed
# @return an integer
def removeElement(self, A, elem):
pt = 0
for i in range(len(A)):
if A[i] != elem:
A[pt] = A[i]
pt += 1
return ptJava版本：
public class Solution
{
public int removeElement(int[] A, int elem) {
int pt = 0;
for (int i = 0; i < A.length; i++)
{
if (A[i] != elem) A[pt++] = A[i];
}
return pt;
}
}本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2015年03月18日，如有侵权请联系 cloudcommunity@tencent.com 删除
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