题目

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.</pre>

解答

这道题的意思是，给我们一个数组，再给一个数字，让我们将数组中不等于这个数字的数移到最前面，并返回长度。

思路
使用双指针的思路，一个指针表示【下一个满足条件的数应该放在什么位置】，另一个指针用来遍历。每遍历到一个满足条件的需要保留的数，就放在第一个指针的下标上。

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