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 = (  + numsSize) * (numsSize+) / ;
     int ret = ;
     int i;
     for( i=; 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=; i<numsSize; i++ )
     {
         if( i != nums[i] )
         {
             return i;
         }
     }
     return numsSize;
 }

 int main()
 {
     int nums[] = { , ,  };
     int numsSize = ;

     int ret = missingNumber2( nums, numsSize );
     printf("%d\n", ret);

     return ;
 }