标签:ice each elf 函数 dice 简单 下标 enum ict
1 """ 2 Given an array of integers, return indices of the two numbers such that they add up to a specific target. 3 4 You may assume that each input would have exactly one solution, and you may not use the same element twice. 5 6 Example: 7 8 Given nums = [2, 7, 11, 15], target = 9, 9 10 Because nums[0] + nums[1] = 2 + 7 = 9, 11 return [0, 1]. 12 13 14 """ 15 16 17 """ 18 第一种是思路简单的方法,先遍历求出目标值,再遍历目标值是否再数组中, 19 判断如果下标相等则继续,最后返回[i,j] 20 """ 21 class Solution1(object): 22 def twoSum(self, nums, target): 23 n = len(nums) 24 for i in range(n): 25 a = target - nums[i] 26 if (a in nums): #这里其实隐含了一层循环 27 j = nums.index(a) 28 if (i==j): 29 continue 30 else: 31 return [i,j] 32 break 33 else: 34 continue 35 """ 36 第二种方法用了dict查找,提高效率 37 enumerate函数用法: 38 seasons = [‘Spring‘, ‘Summer‘, ‘Fall‘] 39 list(enumerate(seasons)) 40 [(0, ‘Spring‘), (1, ‘Summer‘), (2, ‘Fall‘)] 41 """ 42 class Solution2(object): 43 def twoSum(self, nums, target): 44 dict = {} 45 for i, num in enumerate(nums): 46 if target - num in dict: 47 return [dict[target - num], i] 48 else: 49 dict[num] = i 50 nums = [2, 7, 11, 15] 51 target = 9 52 result1 = Solution1() 53 result2 = Solution2() 54 print(result1.twoSum(nums, target)) 55 print(result1.twoSum(nums, target))
标签:ice each elf 函数 dice 简单 下标 enum ict
原文地址:https://www.cnblogs.com/yawenw/p/12250459.html