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].
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
num_maps = {}
for index, value in enumerate(nums):
complement = target - value
complement_index = num_maps.get(complement)
if complement_index is not None:
return [complement_index, index]
num_maps.update({value: index})
return []
思路：
其實很簡單岁诉，題目已經假定肯定有匹配數字罢防，所以不用考慮其它情況。
直接搞個dict唉侄，每次遍歷一個數就把它和它的索引存在里面，然后每次都檢查補數在不在dict里面：
如果在野建，返回補數索引和本次遍歷的數字索引属划。
Complexity Analysis:
Time complexity : O(n). We traverse the list containing n elements only once. Each look up in the table costs only O(1) time.
Space complexity : O(n). The extra space required depends on the number of items stored in the hash table, which stores at most n elements.
