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.

Solution:

public class Solution {
    public int removeElement(int[] nums, int val) {
        int size=0,length=nums.length;
        for(int i=0;i<length;i++){
            if(nums[i]!=val)nums[size++]=nums[i];
        }
        return size;
    }
}

C语言源代码（2ms）：

int removeElement(int* nums, int numsSize, int val) {
    int size=0,i;
    for(i=0;i<numsSize;i++){
        if(nums[i]!=val)nums[size++]=nums[i];
    }
    return size;
}

C++源代码（5ms）：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int size=0,length=nums.size();
        for(int i=0;i<length;i++){
            if(nums[i]!=val)nums[size++]=nums[i];
        }
        return size;
    }
};

Python源代码（64ms）：

class Solution:
    # @param {integer[]} nums
    # @param {integer} val
    # @return {integer}
    def removeElement(self, nums, val):
        size=0;length=len(nums)
        for i in range(length):
            if nums[i]!=val:nums[size]=nums[i];size+=1
        return size