给定一个数字数组nums和一个值val，原地移除所有数值与val相等的元素并返回新长度。

不要分配额外的内存空间给另外的数组，你必须通过原地修改输入的数组以O(1)的额外内存完成这件事。

元素的顺序可以被改变。这不会影响那些你留在新长度后面的元素。

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.

解法1:

因为最终要返回新数组的长度，而新长度后面的元素不会影响结果，所以可以把需要移除的元素交换到数组的末尾，然后返回前面一段（非目标元素）的长度，避免增删数组元素。双指针解法：

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function(nums, val) {
    if (!nums.length) return 0;
        
    let forward = 0;
    let backward = nums.length - 1;
    while(true) {
      while(nums[forward] != val) {
        if (forward == backward) return forward + 1;
        forward ++;
      }
       
      while(nums[backward] == val) {
        if (forward == backward) return backward;
        backward --;
      }
      nums[forward] = nums[backward];
      nums[backward] = val;
      if (forward + 1 >= backward) break;

    }
    return forward + 1;
};