Leetcode刷题记—— Remove Duplicates from Sorted Array II（已排序数组移除重复元素2）

一、题目叙述：

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.

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排序数组中一个元素最多只能重复2次，将多余的移除

二、解题思路：

Medium题。

思路：

（1）倒着遍历数组，若当前元素与其前前一个元素的值不同，计数加1；否则，将后续元素前移。

（2）注意当数组元素小于等于2的情况。

三、源码：

import java.util.Arrays;public class Solution  {  public int removeDuplicates(int[] nums){  if (nums.length <= 2) return nums.length;      int count = 2;      for (int i = nums.length - 1; i >= 2; i--)      {      if (nums[i] != nums[i - 2])      count ++;      else       for (int j = i; j < nums.length - 1; j ++)      nums[j] = nums[j + 1];      }      System.out.println(Arrays.toString(nums));      return count;          }    public static void main(String args[])          {                                  Solution solution = new Solution();              int[] nums = {1, 1, 1, 2, 2, 3};      System.out.println(solution.removeDuplicates(nums));     }      }