80. Remove Duplicates from Sorted Array II
Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Example 1:
Given nums = [1,1,1,2,2,3], Your function should return length =5
, with the first five elements ofnums
being1, 1, 2, 2
and 3 respectively. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,1,2,3,3], Your function should return length =7
, with the first seven elements ofnums
being modified to0
, 0, 1, 1, 2, 3 and 3 respectively. It doesn't matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy) int len = removeDuplicates(nums); // any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) { print(nums[i]); }
解法1:
int removeDuplicates(vector<int>& nums) { if (nums.size() == 0) return 0; int index = 0; int count = 0; int j = 0; for (int i = 0; i < nums.size();){ while (++i<nums.size() && nums[i] == nums[i - 1]); //此时i为分割点的后一个数 count = i - index; //前一类数的个数 index = i; if (count >= 2){ fill(nums.begin() + j, nums.begin() + j + 2, nums[i-1]); j += 2; } else{ nums[j] = nums[i - 1]; j++; } } return j; }
解法2:
int removeDuplicates(vector<int>& nums) { if(nums.size()<=2) return nums.size(); int index = 2; for(int i=2; i<nums.size(); i++){ if(nums[i]!=nums[index-2]) nums[index++] =nums[i]; } return index; }