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, and you may not use the same element twice.

Example:

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

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

任務:給定一個數組，和一個目標的數值，給出相加可以得到目標數值的下標。

暴力搜尋

複雜度\(O(n^2)\)

for i in range(len(nums)):
    for j in range(i+1,len(nums)):
        if nums[i]+nums[j]==target
        	return [i,j]

利用hash表

建立一個雜湊表儲存了數值和下標。複雜度可以為\(O(n)\)

def twosum_hashmap(nums, target):
	'''
	input:
	nums	- list of numbers to calc
	target 	- target value
	output:
	[a,b]	- index of required numbers
	'''
	dataset = {}
	for index in range(len(nums)):
		dataset[(nums[index])] = index
	for value in nums:
		if (target - value) in dataset and \
		(value) in dataset \
		and dataset[(value)] != dataset[(target-value)]:
			return [dataset[(value)], dataset[(target-value)]]
	return []
 

one-pass

def twosum_hashmap_onepass(nums,target):
	dataset={}
	for index in range(len(nums)):
		if (target - nums[index]) in dataset:
			return [index, dataset[(target-nums[index])]]
		else:
			dataset[(nums[index])] = index