Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
给定一个有序的整数数列，每个数最多只能出现两次，然后去除多余的数，把剩下的数排在原数组的前面，并返回新数列的长度
这虽然是个中等题，不过难度不大，唯一需要多考虑的就是在原数组前面排列新的数组
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.
My Solution
(Java) Version 1 Time: 1ms:
做出的方法还是各种立flag,，没什么多解释的，就是遍历一遍，然后跳过重复多的，然后符合标准的就放到前面去，甚至都不用交换
public class Solution {
public int removeDuplicates(int[] nums) {
if (nums.length == 0)
return 0;
int count = 1, tempcount = 1, temp = nums[0], index = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] == temp) {
if (tempcount < 2) {
tempcount++;
nums[index] = nums[i];
index++;
count++;
}
} else {
temp = nums[i];
tempcount = 1;
nums[index] = nums[i];
index++;
count++;
}
}
return count;
}
}
(Java) Version 2 Time: 0ms (By StefanPochmann):
别人家的姿势水平就是高很多，能如此简单地解答，关键在于题目中给出的是有序数列，所以后面的数一定是比前面的大，或者说不同，就不用考虑后面还会有和前面相同数的情况
public int removeDuplicates(int[] nums) {
int i = 0;
for (int n : nums)
if (i < 2 || n > nums[i-2])
nums[i++] = n;
return i;
}
