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.
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 a modification to the input array will be known to the caller.
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]); }
Example 1:
Input: nums = [1,1,1,2,2,3] Output: 5, nums = [1,1,2,2,3] Explanation: Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It doesn't matter what you leave beyond the returned length. Example 2:
Input: nums = [0,0,1,1,1,1,2,3,3] Output: 7, nums = [0,0,1,1,2,3,3] Explanation: Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively. It doesn't matter what values are set beyond the returned length.
Constraints:
0 <= nums.length <= 3 * 104 -104 <= nums[i] <= 104 nums is sorted in ascending order.
public class Solution {
public int RemoveDuplicates(int[] nums) {
int index=0;
for(int i=0;i<nums.Length;i++){
if(index<2 || nums[i] > nums[index-2]){
nums[index++]=nums[i];
}
}
return index;
}
}Time Complexity: O(n)
Space Complexity: O(1)
