LeetCode 80. Remove Duplicates from Sorted Array II

问题描述

Follow up for “Remove Duplicates”:
What if duplicates are allowed at most twice?

For example,
Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn’t matter what you leave beyond the new length.

问题分析

给定一个有序数组。将这个数组中出现超过了两次的数给去掉。返回的结果是当前数组中还剩余的数的个数。
这个题目是使用一个index指针来指定当前处理的位置，如果nums[i] ！= nums[index-2]的话，那么就将nums[index] = nums[i].否者说明nums[i]是需要被排除的值。

代码实现

public int removeDuplicates(int[] nums) {          if (nums.length < 3) {            return nums.length;        }        int index = 2;        for (int i = 2; i < nums.length; i++) {            if (nums[i] != nums[index - 2]) {                nums[index++] = nums[i];            }        }        return index;     }