# 【LeetCode】 27. Remove Element

## Description
> Given an array nums and a value val, 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.

> 給一陣列和一數字val，請移除陣列中所有和val相等的元素，並回傳新的長度。
> 請不要分配額外的空間給一個新的陣列，你應該直接修改輸入提供的陣列。你只能用O(1)的空間。
> 陣列的元素可以改變，你移除了什麼並不影響回傳的長度。

## Example:

Example 1:

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

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

*It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

*It doesn't matter what values are set beyond the returned length.

## Solution

* 掃過一次陣列，相同就移除，複雜度為`O(n)`。
* 注意`vector.size`的提取對時間的影響，你可以先把它放到一個空間裡儲存。

### Code

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

###### tags: `LeetCode` `C++`