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80. Remove Duplicates from Sorted Array II

中文文档

Description

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:

  • 1 <= nums.length <= 3 * 104
  • -104 <= nums[i] <= 104
  • nums is sorted in ascending order.

Solutions

Python3

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        n = len(nums)
        cnt, cur = 0, 1
        for i in range(1, n):
            if nums[i] == nums[i - 1]:
                cnt += 1
            else:
                cnt = 0
            if cnt < 2:
                nums[cur] = nums[i]
                cur += 1
        return cur

Java

class Solution {
    public int removeDuplicates(int[] nums) {
        int cnt = 0, cur = 1;
        for (int i = 1; i < nums.length; ++i) {
            if (nums[i] == nums[i - 1]) ++cnt;
            else cnt = 0;
            if (cnt < 2) nums[cur++] = nums[i];
        }
        return cur;
    }
}

C++

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = nums.size();
        int cnt = 0, cur = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1]) ++cnt;
            else cnt = 0;
            if (cnt < 2) nums[cur++] = nums[i];
        }
        return cur;
    }
};

C#

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        int cnt = 0, cur = 1;
        for (int i = 1; i < nums.Length; ++i)
        {
            if (nums[i] == nums[i - 1]) ++cnt;
            else cnt = 0;
            if (cnt < 2) nums[cur++] = nums[i];
        }
        return cur;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function (nums) {
    if (nums.length == 0) return 0;
    let len = nums.length;
    let j = 0;
    for (let i = 0; i < nums.length - 1; i++) {
        if (nums[i] != nums[i - 1] || nums[i] != nums[i + 1]) {
            nums[j++] = nums[i];
        }
    }
    nums[j] = nums[len - 1];
    return j + 1;
};

Go

func removeDuplicates(nums []int) int {
	i := 0
	for _, num := range nums {
		if i < 2 || num != nums[i-2] {
			nums[i] = num
			i++
		}
	}
	return i
}

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