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