---
description: 善用 nums[pointer] = nums[i] 改變本來陣列，得到不重覆值數量
---

# #26 Remove Duplicates from Sorted Array (有圖)

[LeetCode](https://leetcode.com/problems/remove-duplicates-from-sorted-array/)

```
Given a sorted array nums, 
remove the duplicates in-place such that each element appear only once 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.

output: 過濾掉重覆值然後回傳 length
note: 不要用到額外空間，不會管是劉哪一個eg [1, 1, 1] 你可以濾掉任何一個不用照順序
```

```
Example 1:

Given nums = [1,1,2],

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:

Given nums = [0,0,1,1,1,2,2,3,3,4],

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.
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 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]);
}

/**
 * @param {number[]} nums
 * @return {number}
 */

var removeDuplicates = function(nums) {}
```

### 怎麼解？

他說不要用額外空間所以就不能用 set 了，不然我第一個一定想到 set．題目說已經排序過，所以我就會用 pointer 跟下一個比，nums 關注在記錄不重覆值

![](https://1787585077-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LqirD3iAZIDk4oDCfwS%2F-Lr1BxeeX-sdo0uFsUfj%2F-Lr1KnE9_AVsSMdg_O6Y%2F1.jpg?alt=media\&token=0c6dc2ad-72c8-40e8-ac05-aa2170107fd7)

![](https://1787585077-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LqirD3iAZIDk4oDCfwS%2F-Lr1BxeeX-sdo0uFsUfj%2F-Lr1Kpdk8U-XqAyqbaJV%2F2.jpg?alt=media\&token=5bc3c981-9cf9-47b8-b7dc-7e1d204e7450)

```
 if( nums[pointer] !== nums[i]){
            pointer ++;
            nums[pointer] = nums[i]
        }
```

最後再回傳 pointer 多少就是不重覆值數量了

```
var removeDuplicates = function(nums) {
    const len = nums.length;
    let pointer = 0;
    for(let i = 1; i< len; i++){
        // 如果跟下一個不一樣
        if( nums[pointer] !== nums[i]){
            pointer ++;
            nums[pointer] = nums[i]
        }
    }
    return pointer + 1;
};
```

### 學到什麼?

這題其實蠻幫助思考的，善用 nums\[pointer] = nums\[i] 改變本來陣列
