问题描述：

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.

分析：

和26题要求一样，都不能分配额外的空间出来。和26题的官方答案一个思路，由删除未知个数的重复元素变为删除一个重复元素。同样是两个指针，一个从0出发，存储非指定值，另一个也是从0出发，寻找与指定值不同的值赋给第一个指针。

代码如下：

  public int removeElement(int[] nums, int val) {
        if(nums.length == 0) return 0;
        int i = 0;
        for(int j=0; j<nums.length; j++){
            if(nums[j] != val){
                nums[i++] = nums[j];
            }
        }
        return i;
    }