[LeetCode] 80. Remove Duplicates from Sorted Array II

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

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 k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

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

删除排序数组中的重复项 II。

给你一个有序数组 nums ，请你 原地 删除重复出现的元素，使得出现次数超过两次的元素只出现两次 ，返回删除后数组的新长度。
不要使用额外的数组空间，你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/remove-duplicates-from-sorted-array-ii
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给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素最多出现两次，返回移除后数组的新长度。不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。

思路跟版本一差不多，也是双指针。这里我创建一个 j 指针，从 index = 2 开始，因为前两个位置不需要看了。注意 for 循环里的 i 也是从 2 开始，我们比较的是 nums[i] 与 nums[j - 2] 是否相同，意思是只要这两个数字不同，说明某个数字出现的次数没有超过两次，否则下标为 j, j - 1, j - 2 上的数字应该是相同的。

时间O(n)

空间O(1)

Java实现

class Solution {
    public int removeDuplicates(int[] nums) {
        // corner case
        if (nums.length <= 2) {
            return nums.length;
        }
        
        // normal case
        int j = 2;
        for (int i = 2; i < nums.length; i++) {
            if (nums[i] != nums[j - 2]) {
                nums[j] = nums[i];
                j++;
            }
        }
        return j;
    }
}

 

JavaScript实现

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function (nums) {
    // corner case
    if (nums.length <= 2) {
        return nums.length;
    }

    // normal case
    let j = 2;
    for (let i = 2; i < nums.length; i++) {
        if (nums[i] !== nums[j - 2]) {
            nums[j] = nums[i];
            j++;
        }
    }
    return j;
};

 

LeetCode 题目总结