1. Two Sum

题目：

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

想法

对nums建立hash表，再遍历nums，看hash表中有没有相对应的关键字加上num会等与target

代码

int* twoSum(int* nums, int numsSize, int target) {
    int min = nums[ 0 ], max = nums[ 0 ];
    int *arr = malloc( sizeof( int ) * 2 );
    
    for( int i = 1; i < numsSize; i++ )
    {
        if( min > nums[ i ] )
        {
            min = nums[ i ];
        }
        
        if( max < nums[ i ] )
        {
            max = nums[ i ];
        }
    }
    
    int *hash = malloc( sizeof( int ) * ( max - min + 1 ) );
    
    for( int i = 0; i < max - min + 1; i++ )
    {
        hash[ i ] = -1;
    }
    
    for( int i = 0; i < numsSize; i++ )
    {
        int key = target - nums[ i ] - min;
      
        if( key < 0 || key >= max - min + 1)
        {
            continue;
        }
        
        if( hash[  key ] >= 0  )
        {
          
            
            arr[ 0 ] = i;
            arr[ 1 ] = hash[ key ];
            
            free( hash );
            return arr;
        }
        hash[ nums[ i ] - min ] = i;
    }
    
    free( hash );
    
    return arr;
}