原题传送门

1. 题目描述

2. Solution 1

1、思路分析
Question wants us to return the length of new array after removing duplicates and that we don't care about what we leave beyond new length , hence we can use i to keep track of the position and update the array.
2、代码实现

package Q0099.Q0080RemoveDuplicatesFromSortedArrayII;  public class Solution {     /*        Q026 的续集，Remove Duplicates from Sorted Array II (allow duplicates up to 2):      */     public int removeDuplicates(int[] nums) {         int i = 0;         for (int n : nums) {             if (i < 2 || n > nums[i - 2]) nums[i++] = n;         }         return i;     } } 

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(1)