1. 问题描述

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.

2. 解法

2.1 暴力搜索

class Solution:
    def twoSum(self, nums, target):
        L = []
        for i in range(0, len(nums)):
            for j in range(i, len(nums)):
                if(i!=j and nums[i] + nums[j] == target and ((target%2 == 0 and nums[i]%2==nums[j]%2) or (target % 2 !=0 and (nums[i]%2)^(nums[j]%2)))):
                    L.append(i)
                    L.append(j)
                    return L

对奇偶和两个数满足的条件进行了筛选，时间复杂度仍为

2.2 Hash Table查找(One pass)

2.2.1 思路

依次遍历List，游标为index。将target-List[index]的值存在字典中，继续查找List，如果随后List中的元素出现在字典的Key中，说明找到了List的当前元素和之前遍历过的元素的配对，进而返回对应的列表[dict[nums[index]], index]；Two pass的做法是先构造一个完整的字典，再进行对应的Hash Table的查找，时空复杂度依旧不变。

import time
class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for index in range(len(nums)):
            if nums[index] in dict:               
                return [dict[nums[index]], index]
            else:
                dict[target-nums[index]] = index
        print(dict)

2.2.2 复杂度分析

只对列表遍历了一遍，时间复杂度为， 同时构造了一个字典（Hash Table），空间复杂度是， 属于空间换时间的做法。