Leetcode 27. Remove Element（too easy）

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() == 0){
            return 0;
        }
        int newindex = 0;
        for (int i = 0; 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。