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.
用map存每个数及其出现次数,出现一次的,被推上新数组一次,多次的,被推上新数组两次。
在原数组上进行改变。用两个指针,一个i用来指数组改变到的位置,一个j用来指遍历到的位置。遍历时,判断nums[j]和其下一个数nums[j+1]是否相同,如果相同的话,nums[i]和nums[i+1]都等于这个数,然后,将指针j移到其后面第一个与nums[i]存的数不一样的数,继续遍历,而i也要i+=2,指向存好的两个数之后。如果nums[j]和nums[j+1]不同,则存一次,两个指针同时向后一位。
class Solution { public: int removeDuplicates(vector<int>& nums) { int n = nums.size(); if (n == 0) return 0; int i = 0, j = 0; while(i < n && j < n){ if (j+1 < n && nums[j] == nums[j+1]){ nums[i] = nums[j]; nums[i+1] = nums[j]; while(nums[i] == nums[j]){ j++; } i+=2; } else{ nums[i] = nums[j]; i++; j++; } } return i; } };