1. 题目

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 in place with constant memory.

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

Example:
Given input array nums = [3,2,2,3], val = 3

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

2. 思路

遍历，遇到待删除的元素，将当前的末尾移到当前位置，继续处理当前位置。
当前末尾也可以持续前进直到遇到不是待删除元素，或者是全部处理完了。

3. 代码

耗时：3ms

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int sz = nums.size();
        if (sz == 0) return 0;
        int n = 0;
        int k = 0;
        for (int i = 0; i < sz - k; i++) {
            if (nums[i] == val) {
                int pos = sz - k - 1;
                if (pos <= i) {
                    return n;
                }
                nums[i] = nums[pos];
                i--;
                k++;
            } else {
                n++;
            }
        }
        return n;
    }
};