# 思路
类似第26题, 用count记录非val的个数,从前往后遍历,如果值为val则跳过,否则令nums[count并=nums[i]并自增count
# C++
``` C++
class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int count = 0;
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] != val) nums[count++] = nums[i]; 
        }
        return count;
    }
};
```