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.

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.

思路

用快慢指针的方法遍历数组, 快指针从0-i遍历, 慢指针记录快指针所有不等于val的数值

复杂度

时间O(n) 空间O(1)

代码

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