Two Sum

找了个leeCode最简单的题做，结果发现要做到O(n)还是好麻烦……

#include <stdio.h>
#include <stdlib.h>

typedef struct TANode
{
	int pos;
	struct TANode *next;
}TA,*TAP;

int* twoSum(int* nums, int numsSize, int target) {
	int max=nums[0],i,offset,min=nums[0];
	for(i=1;i<numsSize;i++)
	{
		if(nums[i]>max)
			max=nums[i];
		if(nums[i]<min)
			min=nums[i];
	}
	if(min<0)
		min=-min;
	int *r=(int*)malloc(2*sizeof(int));
	TA *ta=(TA *)malloc((max+min+1)*sizeof(TA));
	for(i=0;i<(max+min+1);i++)
	{
		ta[i].pos=-1;
		ta[i].next=NULL;
	}
	for(i=0;i<numsSize;i++)
	{
		if(ta[nums[i]+min].pos==-1)
		{
			ta[nums[i]+min].pos=i;
		}
		else
		{
			TAP t=(TAP)malloc(sizeof(TA));
			t->pos=i;
			t->next=NULL;
			ta[nums[i]+min].next=t;
		}
	}
	for(i=0;i<numsSize;i++)
	{
		int x=nums[i],y;
		if(target-x+min>=0 && ta[target-x+min].pos != -1)
		{
			if(i<ta[target-x+min].pos)
			{
				r[0]=i;
				r[1]=ta[target-x+min].pos;
				return r;
			}
			else if(i>ta[target-x+min].pos)
			{
				r[0]=ta[target-x+min].pos;
				r[1]=i;
				return r;
			}
			else
			{
				TAP t=&ta[target-x+min];
				if(t->pos==i)
				{
					t=t->next;				
				}
				if(t)
				{
					r[0]=i;
					r[1]=t->pos;
					return r;
				}

			}
		}
	}
	return r;
}

void printA(int *a,int len)
{
	for(int i=0;i<len;i++)
	{
		printf("%d ",a[i]);
	}
}

void main()
{
	int nums[3]={3,3,6};
	int *r=twoSum(nums,3,6);
	printA(r,2);
	getchar();
}