[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，所以思路是做两个指针，一个从nums[1]开始指向目前遍历到的数字，一个res指针指向0。如果当前数字nums[i] != nums[i - 1]，则将nums[i]放入res所在的位置。

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public int removeDuplicates(int[] nums) {
 3         // corner case
 4         if (nums == null || nums.length == 0) {
 5             return 0;
 6         }
 7 
 8         // normal case
 9         int res = 1;
10         for (int i = 1; i < nums.length; i++) {
11             if (nums[i - 1] != nums[i]) {
12                 nums[res] = nums[i];
13                 res++;
14             }
15         }
16         return res;
17     }
18 }

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var removeDuplicates = function(nums) {
 6     // corner case
 7     if (!nums || nums.length === 0) {
 8         return 0;
 9     }
10 
11     // normal case
12     let res = 1;
13     for (let i = 1; i < nums.length; i++) {
14         if (nums[i] !== nums[i - 1]) {
15             nums[res] = nums[i];
16             res++;
17         }
18     }
19     return res;
20 };

LeetCode 题目总结