1. Two Sum【数组|哈希表】

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.

版本1：O(n^2)  【暴力 原始版本】O（1）

1. classSolution(object):
2. def twoSum(self, nums, target):
3. length = len(nums)
4. for i in range(length):
5. for j in range(i+1,length)://游标后移
6. if nums[i]+ nums[j]== target:
7. return[i,j]

版本2：O(n^2)  【暴力 枚举函数增强】O（1）

1. classSolution(object):
2. def twoSum(self, nums, target):
3. for index , item in enumerate(nums):
4. for past_index , past_item in enumerate(nums[index+1:],index+1):
5. if item + past_item == target:
6. return[index, past_index]

版本3：O(n)    【双程hash table】O（n）

1. classSolution(object):
2. def twoSum(self, nums, target):
3. dd = dict()
4. for index , value in enumerate(nums):
5. dd[value]= index
6. for index , value in enumerate(nums):
7. tt = target - value
8. if dd.has_key(tt)and dd[tt]!= index:
9. return[index , dd[tt]]

版本4：O(n)    【单程hash table】O（n）

Python

1. classSolution(object):
2. def twoSum(self, nums, target):
3. dd = dict()
4. for index , value in enumerate(nums):
5. tt = target - value
6. if dd.has_key(tt)and dd[tt]!= index:
7. return[dd[tt],index]
8. else:
9. dd[value]= index

Java

1. public int[] twoSum(int[] nums, int target){
2. Map<Integer,Integer> map = new HashMap<Integer,Integer>();
3. for( int i=0;i<nums.length;i++){
4. if(!map.containsKey(target - nums[i])){
5. map.put(nums[i], i );
6. }else{
7. int[] rs ={ map.get(target-nums[i]), i };
8. return rs;
9. }
10. }
11. return nums;
12. }

1.enumerate内置函数的使用（枚举函数）---用于既要遍历索引又要遍历元素；可以接收第二个参数用于指定索引起始值。