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
題解：
就是設置一個limit，設置當前count為1，用來返回結果的index為1.
每次在循環里(從1開始)嘗試更新count， 假如nums[i] = nums[i - 1]則count++，否則count = 1
在count <= limit的條件下，我們可以更新num[index++] = nums[i]。
最后返回index。index從1開始。
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;
}
}
