标签：integer elf == 索引 twosum ret ict bsp rate
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].
Solution_1
简单的for语句
class Solution(object): def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ for index1, num1 in enumerate(nums): for index2, num2 in enumerate(nums): if index1 != index2: if num1 + num2 == target: return [index1, index2]
Solution_2
用一个字典来表示每个索引和target之间的差。然后判断列表中的元素是否在字典的键值中
class Solution(object): def twoSum(self, nums, target): if len(nums) <= 1: return False buff_dict = {} for i in range(len(nums)): if nums[i] in buff_dict: return [buff_dict[nums[i]], i] else: buff_dict[target - nums[i]] = i
标签：integer elf == 索引 twosum ret ict bsp rate
原文地址：https://www.cnblogs.com/francischeng/p/9747057.html
