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题目：
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.
思路：记录每个数字的次数，并且维护新数组的位置。
public class Solution { public int removeDuplicates(int[] nums) { if(nums.length == 0) return 0; int walker = 1; //每个的数量 int count = 1; //总类数量 int loc = 1; //定位 for(int i = 1 ; i < nums.length ; i++){ if(nums[i] == nums[i-1]){ if(walker < 2){ walker++; count++; }else{ //出现次数大于2次 walker++; continue; } }else{ walker = 1; //出现新数字 count++; } nums[loc] = nums[i]; loc++; } return count; } }
[LeetCode-JAVA] Remove Duplicates from Sorted Array II
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原文地址：http://www.cnblogs.com/TinyBobo/p/4510441.html
