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.
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题意：一个有序数组，每个元素最多出现两次，如果大于两次，把多的去掉，返回去掉后的数组长度len，以及要求数组前len是去掉那些元素之后的元素。//语死早。。看原题好了。。
思路：排序了还不是随便搞？ 没要求空间再开一个标记空间。。。O(1)空间的话。。就乱搞一下？
1/* ***********************************************
2Author :111qqz
3Created Time :2017年04月05日 星期三 21时25分48秒
4File Name :80.cpp
5************************************************ */
6class Solution {
7
8public:
9
10 int removeDuplicates(vector<int>& nums) {
11 int siz = nums.size();
12 if (siz==0) return 0;
13 int p = 0;
14 int magic = -324784312;
15 for ( int i = 0 ; i < siz-1 ; i++)
16 {
17 if (nums[i]==nums[i+1]) p++;
18 else p = 0;
19 if (p>=2) nums[i-1] = magic;
20 }
21 int cur = 0;
22 for ( int i = 0 ; i < siz ; i++)
23 {
24 if (nums[i]!=magic) nums[cur++] = nums[i];
25 }
26 return cur;
27
28 }
29
30};