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

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

class Solution {

public:

    int removeElement(vector<int>& nums, int val) {

        if (nums.size() == ){

            return ;

        }

        int newindex = ;

        for (int i = ; i < nums.size(); i++){

            if (nums[i] != val){

                nums[newindex] = nums[i];

                newindex++;

            }

        }

        return newindex;

    }

};

思路很简单，o（n）的时间复杂度，没有开辟另外的空间，感觉之前做过一个类似的问题。