[leetcode]27. Remove Element删除元素

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.

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.

题意：

给定某个值，要求删除数组中所有等于该值的元素。

思路：

j

以[3, 2, 2, 3]， value = 3 为例

i

指针i遍历数组，遇到非value的值，送到指针j那里去

指针j从0开始，接收所有指针i送来的非value值

扫完一遍后

指针j所指的位置就是新“数组”的长度

代码：

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