[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。

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思路：

指针i遍历数组，遇到非value的值，送到指针start那里去

指针start从0开始，接收所有指针i送来的非value值

扫完一遍后

指针start所指的位置就是新“数组”的长度

代码：

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