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;}