[LeetCode] 26. Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns 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 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]);
}

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4]
Explanation: Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 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.

删除排序数组中的重复项。

给你一个有序数组 nums ，请你 原地 删除重复出现的元素，使每个元素 只出现一次 ，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.com/problems/remove-duplicates-from-sorted-array
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既然题目要求了不能用额外空间，这里考虑用双指针的思路做。对于每个位置上的数字 nums[i]，如果他与之前一个数字不一样，就可以被加入结果。如果这个数字之后出现了多次，那么也他也只有只一次被加入结果。

时间O(n)

空间O(1)

Java实现

class Solution {
    public int removeDuplicates(int[] nums) {
        // corner case
        if (nums == null || nums.length == 0) {
            return 0;
        }

        // normal case
        int j = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[i - 1]) {
                nums[j] = nums[i];
                j++;
            }
        }
        return j;
    }
}

 

JavaScript实现

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    // corner case
    if (!nums || nums.length === 0) {
        return 0;
    }

    // normal case
    let res = 1;
    for (let i = 1; i < nums.length; i++) {
        if (nums[i] !== nums[i - 1]) {
            nums[res] = nums[i];
            res++;
        }
    }
    return res;
};

 

LeetCode 题目总结