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.

感觉这道题的oa有点问题。。。先这样吧

public class Solution {
public int removeDuplicates(int[] nums) { int res=0;
int len=nums.length;
int set=0; for(int i=0; i<len; i++){
if(i==0){
set++;
res++;
}
else{
if(nums[i]==nums[i-1]){
set++;
if(set<=2){
res++;
}
}
else{
set=1;
res++;
}
} }
return res;
}
}

要in place 改变原数组:

public class Solution {
public int removeDuplicates(int[] nums) {
if(nums==null || nums.length==0){
return 0;
}
int startPosition=0;
boolean isRepeated=false;
for(int i=1; i<nums.length; i++){
if(nums[i] != nums[startPosition]){
isRepeated=false;
startPosition++;
nums[startPosition] = nums[i];
}
else{
if(isRepeated == false){
startPosition++;
nums[startPosition]=nums[i];
isRepeated=true;
}
}
}
return startPosition+1; }
}

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