---
title: 'LeetCode 26. Remove Duplicates from Sorted Array'
disqus: hackmd
---

# LeetCode 26. Remove Duplicates from Sorted Array

## Description

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. 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.

## Example

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.

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 k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

## Constraints

1 <= nums.length <= 3 * 10^4^
-100 <= nums[i] <= 100
nums is sorted in non-decreasing order.

## Answer

目標是將所有重複項都拿掉，每個數值都是唯一項，設cnt為唯一項答案，i為往下找數值不同項，若有相同項出現則cnt != i，此時才需要將第i項數值更新到第cnt項數值，所以當我們檢查到項目數值不同的時候還要檢查cnt是否等於i，才決定要不要更新。

int removeDuplicates(int* nums, int numsSize){
    int i = 0, cnt = 1;
    for(i = 1; i < numsSize; i++){
        if(nums[i-1] != nums[i]){
            if(i != cnt){nums[cnt] = nums[i];}
            cnt++;
        }
    }
    return cnt;
}

## Link
https://leetcode.com/problems/palindrome-number/
###### tags: `Leetcode`