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.
一刷
题解：
就是设置一个limit，设置当前count为1，用来返回结果的index为1.
每次在循环里(从1开始)尝试更新count， 假如nums[i] = nums[i - 1]则count++，否则count = 1
在count <= limit的条件下，我们可以更新num[index++] = nums[i]。
最后返回index。index从1开始。
public class Solution {
public int removeDuplicates(int[] nums) {
if(nums == null || nums.length == 0) return 0;
int limit = 2, count = 1, lo = 1;
for(int i=1; i<nums.length; i++){
count = (nums[i] == nums[i-1]) ? count+1:1;
if(count<=limit){
nums[lo] = nums[i];
lo++;
}
}
return lo;
}
}
