【LeetCode】1. Two Sum

1. Two Sum
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.

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

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


看到这个题首先想到的是保存二重循环，先保存一个值，然后去找去找target-val

 1 int* twoSum(int* nums, int numsSize, int target) {
 2     int  i,j;
 3     int bi=-1,bj=-1,tmp;
 4     for(i=0;i<numsSize-1;i++)
 5     {
 6         bi=i;
 7         tmp=target-nums[i];
 8         for(j=i+1;j<numsSize;j++)
 9         {
10             if(tmp==nums[j])
11                 {
12                     bj=j;
13                     break;
14                 }
15         }
16         if(bj!=-1)
17             break;
18     }
19     int *ptmp=malloc(sizeof(int)*2);
20     ptmp[0]=bi;
21     ptmp[1]=bj;
22     return ptmp;
23 }

 方法二：O(n)

每遍历一个数字val时候，首先查找在hash表中有没有target-val。

1.如果有的话，break，然后找到target-val的位置，返回；

2.如果没有找到target-val的话，在hash中记录此数字。

 1 /**
 2  * Note: The returned array must be malloced, assume caller calls free().
 3  */
 4 int* twoSum(int* nums, int numsSize, int target) {
 5     int  i,j;
 6     int hash[10000]={0};　　　　//存储正值
 7     int hashp[10000]={0};　　　　//存储负值
 8     int couple;
 9     for(i=0;i<numsSize;i++)
10     {
11         couple=target-nums[i];
12         if(couple>0)
13             {
14                 if(hash[couple>>5]&(1<<(couple&0x1f)))
15                 break;
16             }
17         else
18             {
19                 if(hashp[(-couple)>>5]&(1<<((-couple)&0x1f)))
20                 break;
21             }
22         if(nums[i]>0)
23             hash[nums[i]>>5]|=(1<<(nums[i]&0x1f));
24         else
25             hashp[(-nums[i])>>5]|=(1<<((-nums[i])&0x1f));            
26     }
27     for(j=0;j<i;j++)
28         if(couple==nums[j])
29             break;
30     int*n=malloc(sizeof(int)*2);
31     n[0]=j;
32     n[1]=i;
33     return n;
34 }

 python版：

 1 class Solution:
 2     def twoSum(self, nums, target):
 3         """
 4         :type nums: List[int]
 5         :type target: int
 6         :rtype: List[int]
 7         """
 8         flag = set()
 9         res = []
10         for i, item in enumerate(nums):
11             if target-item in flag:
12                 res.append(i)
13                 break
14             else:
15                 flag.add(item)
16         if res:
17             res.append(nums.index(target-item))
18         return res
19

 

 1 class Solution:
 2     def twoSum(self, nums, target):
 3         if len(nums)<=1:
 4             return False
 5         buff = {}
 6         for i,v in enumerate(nums):
 7             if v in buff:
 8                 return [buff[v], i]
 9             else:
10                 buff[target-v]=i

 

posted @ 2016-12-08 20:56  wilderness  阅读(136)  评论(0)  编辑  收藏  举报