从数组中寻找和的相加数

Two Sum

• 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 1

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

example 2

Given nums = [0,4,3,0], target = 0,Because nums[0] + nums[3] = 0 + 0 = 0,return [0, 3].

example 3

Given nums = [-1,-2,-3,-4,-5], target = -8,Because nums[2] + nums[4] = -3 + (-5) = -8,return [2, 4].

example 4

Given nums = [0,1,2,3,4,5,...,28888,...,65432], target = 65432,you should consider that time and space is limited.

思路

1. 这个题目简单，两个for循环嵌套可以实现，但是考虑到输入数组长度大的情况下，时间复杂度太高, O(n²)
2. 使用索引，遍历一次后记录nums中所有数字出现的下标，用target - nums[i]得到j，然后直接查索引中j的位置
3. 索引可以用list(数组)，但是nums中可能会有大数字，所需的空间太多，并且会有负数，改用dir，key为nums[i]，value为i
4. 同时给出简单版代码（简单版在java，C等语言在leetcode网站可以AC，但python会超时）

代码

优化版

class Solution:    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        idx_set = {}        for i, item in enumerate(nums):            idx = target - nums[i]            if idx in idx_set:                j = idx_set[idx]                if len(j) != 0 and j[0] != i:                    return list([i, j[0]])                if j[0] == i and len(j) > 1:                    return list([i, j[1]])            else:                idx_set[item] = [i]

简单版

class Solution:    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        for i in range(len(nums)):            for j in range(i+1, len(nums)):                if nums[i] + nums[j] == target:                    return list([i, j])

本题以及其它leetcode题目代码github地址: github地址