LeetCode 27. Remove Element

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.

移除数组中和给定值相同的元素，返回新数组的长度。不允许分配新的空间来制造新的数组，数组元素的顺序可以改变，新数组长度外的元素不需要理会。
这题还是用Two Pointers，一个指向数组当前尾元素（end），一个指向当前元素（i），若当前元素不等于给定值，那么i移到下一个元素；若当前元素等于给定值，就把尾元素的值赋给当前元素，并把end指针向前移一位。

int removeElement(vector<int>& nums, int val) {        if (nums.size() == 0) return 0;        int end = nums.size() - 1, i = 0;        while (i <= end)        {            if (nums[i] == val) {                nums[i] = nums[end];                --end;            }            else ++i;        }        return i;    }