LeetCode 268

Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n,
find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity.
Could you implement it using only constant extra space complexity?

/*************************************************************************
    > File Name: LeetCode268.c
    > Author: Juntaran
    > Mail: Jacinthmail@gmail.com
    > Created Time: Tue 10 May 2016 06:19:13 PM CST
 ************************************************************************/

/*************************************************************************

    Missing Number
    
    Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, 
    find the one that is missing from the array.

    For example,
    Given nums = [0, 1, 3] return 2.

    Note:
    Your algorithm should run in linear runtime complexity. 
    Could you implement it using only constant extra space complexity?

 ************************************************************************/
 
#include <stdio.h>

/* 通用方法 */
int missingNumber( int* nums, int numsSize )
{
    int sum = ( 0 + numsSize) * (numsSize+1) / 2;
    int ret = 0;
    int i;
    for( i=0; i<numsSize; i++ )
    {
        ret += nums[i];
    }
    ret = sum - ret;
    return ret;
}

/* 如果数据是从0递增的话可以使用以下方法 */
/* 测试用例包含不是纯从0递增数组，所以此方法LeetCode不通过 */
int missingNumber2( int* nums, int numsSize )
{
    int i;
    
    for( i=0; i<numsSize; i++ )
    {
        if( i != nums[i] )
        {
            return i;
        }
    }
    return numsSize;
}

int main()
{
    int nums[] = { 0, 1, 3 };
    int numsSize = 3;
    
    int ret = missingNumber2( nums, numsSize );
    printf("%d
", ret);

    return 0;
}