LeetCode -- Remove Duplicates From Sorted Array 2
题目描述：
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.
移除数组中的重复元素，返回新长度。
思路：
1.two pointer方法，使用两个指针一前一后（i后，j前），初始化：i=2，j=1
2.i一直往后走，如果nums[i]不等于nums[j]，或nums[i]不等nums[j-1]：nums[j]=nums[i]，j往后走。
3.最后返回j+1
实现代码：
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.
移除数组中的重复元素，返回新长度。
思路：
1.two pointer方法，使用两个指针一前一后（i后，j前），初始化：i=2，j=1
2.i一直往后走，如果nums[i]不等于nums[j]，或nums[i]不等nums[j-1]：nums[j]=nums[i]，j往后走。
3.最后返回j+1
实现代码：
public int RemoveDuplicates(int[] nums)
{
if(nums == null){
return 0;
}
var n = nums.Length;
if(n < 3){
return n;
}
var p = 1;
for(var i = 2;i < n; i++){
if(nums[i] != nums[p] || nums[i] != nums[p-1]){
p ++;
nums[p] = nums[i];
}
}
return (p + 1);
}