内容简介:Given an array of integers, returnYou may assume that each input would have给定一个整数数组和一个目标值,找出数组中和为目标值的
- 英文
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.
- 中文
给定一个整数数组和一个目标值,找出数组中和为目标值的 两个 数。
你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用。
示例
给定 nums = [2, 7, 11, 15], target = 9 因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1]
题解
- 题解 1
两层循环暴力求解,对于列表的每个数,依次使其与其后所有的数单独求和,判断和是否为目标值。
class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ length = len(nums) for i in range(length - 1): for j in range(i + 1, length): # i 之后的数 if nums[i] + nums[j] == target: # 判断是否等于目标值 return i, j
- 题解 2
用一层循环,直接在里面查询 target-nums[i]
是否存在于 nums
列表中,速度比题解 1 快了许多。
class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ for i in range(len(nums)): x = target - nums[i] if x in nums: # 是否有对应的数与 nums[i] 的和为目标值 j = nums.index(x) if i == j: continue else: return i, j # 输出结果
- 题解 3
遍历数组,然后使用字典,键为 target-nums[i]
,值为 i
,依次判断 nums[i]
是否已经在字典中,如果是,则输出结果,即和为目标值的两个数。否则把 target-nums[i]
的值存入字典。
class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ d = {} for i in range(len(nums)): x = target - nums[i] if nums[i] in d: # 判断是否已经在字典中 return d[nums[i]], i # 输出结果 else: d[x] = i # 存入字典
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