---
title: 'LeetCode 268. Missing Number'
disqus: hackmd
---

# LeetCode 268. Missing Number

## Description

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

## Example

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

## Constraints

n == nums.length
1 <= n <= 10^4^
0 <= nums[i] <= n
All the numbers of nums are unique.

## Answer

此題可用for會遞增的概念去當成0~size的疊加，然後減掉nums已有的數，最後仍可找出缺的數值。

int missingNumber(int* nums, int numsSize){
    int i=0,ans=numsSize;
    for(i=0;i<numsSize;i++){
        ans = ans - nums[i] + i;
    }
    return ans;
}

秉持"做兩次XOR會回到原本的數"，一開始就將0~size的數全部XOR起來，然後對nums再做一次XOR，即可找出缺的數。

int missingNumber(int* nums, int numsSize){
    int i = 0,ans = numsSize;
    for(i = 0;i<numsSize;i++){
        ans = ans^i^nums[i];
    }
    return ans;
}

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