首先,需要创建一个函数,它接受两个参数:一个整数数组和一个目标值。该函数将用于查找数组中相加和等于目标值的两个数字。 登录后复制deftwoSum(nums, target): 接下来,需要初始化两个变量登录后复制i和登录后复制j,分别设为 0。这两个变量将用于跟踪数组中相加和等于目标值的两个数字的索引。 登录后复制deftwo...
publicint[]twoSum(int[] nums,inttarget){for(inti =0; i < nums.length; i++) {for(intj = i +1; j < nums.length; j++) {if(nums[j] == target - nums[i]) {returnnewint[] { i, j }; } } }thrownewIllegalArgumentException("No two sum solution"); } 复杂度分析: 时间复杂度...
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. classSolution(object):deftwoSum(self, nums, target):""":type nu...
PYTHON 方法/步骤 1 新建一个PY文档，打开JUPTER NOTEBOOK。2 #Given nums = [2, 7, 11, 15], target = 9nums = [2, 7, 11, 15]target = 9我们要找出列表里面两个相加数为目标的数字。3 nums = [2, 7, 11, 15]target = 9nums2 = numsfor a, j in enumerate(nums): for b, k in ...
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. * 我的理解:
Two Sum 两数之和 (python)题目Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.You may assume that each input would have exactly one solution, and you may not use the same element twice....
Runtime: 848 ms, faster than32.81%ofPython3online submissions forTwo Sum. 仍旧是连一半都没超过啊,娘匹西。 官方方法 Two-Pass Hash Table class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashTable = {} length = len(nums) for i in range(length): hashTabl...
classSolution:deftwoSum(self,nums:List[int],target:int)->List[int]:foriinrange(0,len(nums)):remain=target-nums[i]#print(remain)ifremaininnumsandnums.index(remain)!=i:returni,nums.index(remain) 结果: 3. Hash Table In Python, the hash table we use is the dictionary. ...
class Solution: 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 ...
python代码如下: 1 class solution: 2 def twosum(self, nums, target): 3 for i in range(len(nums)): 4 for j in range(i+1 , len(nums)): 5 if nums[i] + nums[j] == target: 6 return [i, j] 【方法2: one-pass hash table】 如果我们想要降低时间复杂度,该如何解决呢?我们可以...