1. Both Sum "array | hash table"

Given an array of integers, return indices of the both numbers such that they add-to a specific target.

You may assume this each input would has exactly one solution.

Version 1:o (n^2) "Violent original Version" O (1)

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

Version 2:O (n^2) "Brute Force enumeration function enhancement"O (1)

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

Version 3:o (n)"Two-way hash table"O (n)

2. classSolution(object):
4. def twoSum(self, nums, target):
6. dd = dict()
8. for index , value in enumerate(nums):
10. dd[value]= index
12. for index , value in enumerate(nums):
14. tt = target - value
16. if dd.has_key(tt)and dd[tt]!= index:
18. return[index , dd[tt]]

Version 4: O (n)"One way hash table"O (n)
Python

2. classSolution(object):
4. def twoSum(self, nums, target):
6. dd = dict()
8. for index , value in enumerate(nums):
10. tt = target - value
12. if dd.has_key(tt)and dd[tt]!= index:
14. return[dd[tt],index]
16. else:
18. dd[value]= index

Java

2. public int[] twoSum(int[] nums, int target){
4. Map<Integer,Integer> map = new HashMap<Integer,Integer>();
6. for( int i=0;i<nums.length;i++){
8. if(!map.containsKey(target - nums[i])){
10. map.put(nums[i], i );
12. }else{
14. int[] rs ={ map.get(target-nums[i]), i };
16. return rs;
18. }
20. }
22. return nums;
24. }

The use of 1.enumerate built-in functions (enumeration functions)---Used to traverse the index and iterate through the elements; You can receive the second parameter to specify the index starting value.

1. Both Sum "array | hash table"