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;
}
