Follow up for "Remove Duplicates":What if duplicates are allowed at mosttwice?
For example,Given sorted arraynums=[1,1,1,2,2,3],
Your function should return length =5, with the first five elements ofnumsbeing1,1,2,2and3. It doesn’t matter what you leave beyond the new length.
思路：本题允许数组最多有两个重复，，所以需要加一个判断，是否到两个。具体代码如下：
public class Solution {public int removeDuplicates(int[] nums) {if(nums.length <= 2){//长度小于2，直接返回return nums.length;}boolean isTwice = false;//是否两次int len = 0;//最新长度for(int i = 0; i < nums.length; i++){//遍历//不等于最后一个切数字相等if(i != nums.length -1 && nums[i+1] == nums[i]){if(!isTwice){//还没两次isTwice = true;nums[len++] = nums[i]; //添加到数组最前}}else{//不相等isTwice = false;//标记为不是两次nums[len++] = nums[i];//添加最前}}return len;}}
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