这个代码完全不能通过LeetCode的测试。算法复杂度太高 然后我看了别人写的 1classSolution(object):2deftwoSum(self,nums,target):3"""4:type nums: List[int]5:type target: int6:rtype: List[int]7"""8n =len(nums)9result ={}10ifn <= 1:11return
def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerate(nums): if target - num in hashtable: return [hashtable[target - num], i] hashtable[nums[i]] = i return [] 官方给出的答案里,有些函数和语句可能不太了解,这里我说明一下...
其它解法六:LeetCode 中国的普通解法,和解法二类似 class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: n = len(nums) for i in range(n): for j in range(i + 1, n): if nums[i] + nums[j] == target: return [i, j] return [] ## 找不到则返回空...
因为不是计算机专业的,算法基础有点差,所以最近开始在leetcode上刷刷题补一补。 #1 问题链接:twoSum 问题描述: Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to ...
这道题是大名鼎鼎的LeetCode的第一题,也是面试当中非常常见的一道面试题。题目不难,但是对于初学者来说应该还是很有意思,也是一道很适合入门的算法题。 废话不多说,让我们一起来看看题目吧。 Given an array of integers, return indices of the two numbers such that they add up to a specific target. You...
Leetcode c++语言 方法/步骤 1 问题描述：给定一个整数数组，返回两个数字的索引，使它们相加的值等于一个特定的目标值。假设对于每个输入只有一种解决方案，并且您不可以两次同时使用相同的元素。2 问题的示例：给定nums = [2,7,11,15]， target = 9，因为nums[0] + nums[1] = 2 + 7 = 9，返回[0,...
针对2Sum,先对数组排序,然后使用双指针匹配可行解就可以解决,虽然可以考虑使用HashMap加速搜索,但是对于本题使用HashMap的与否的时间复杂度都一样,都是O(nlog(n))。可以参考这个链接: 求和问题总结(leetcode 2Sum, 3Sum, 4Sum, K Sum),写的很清楚。
Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers suchthat they add up to the target, where index1 must be less than index2.Please note that your returned answers (both index1 and...
您必须返回*returnSize中数组中的元素数，因为调用者需要它。(a)Leetcode没有在problem page上说明这个...
leetcode算法—两数之和 Two Sum 关注微信公众号:CodingTechWork,一起学习进步。 题目 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...