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 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标。 你可以假设每种输入只会对应一个答案。但是,你不能重复利用这个数组中同样的元素。 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]
#解法一:求差值,判断差值是否在nums数组里 class Solution: def twoSum(self,nums,target): n = len(nums) for x in range(n): b = target - nums[x] if b in nums: y = nums.index(b) if y != x: return [x,y] nums = [2,7,11,13] target = 9 res = Solution().twoSum(nums,target) print(res) #解法二:求差值、把差值存进字典里作为键、索引作为值,第一次循环理解:d[7]=0 即字典d={7:0},表示为索引0需要数组里值为7的元素配对。 if 判断是否为前面元素所需要配对的值 , 是则返回两个索引值。(补充:nums[x] in d 是判断值是否在字典某个key里面) class Solution: def twoSum(self,nums,target): n = len(nums) d ={} for x in range(n): a = target - nums[x] if nums[x] in d: return d[nums[x]],x else: d[a] = x nums = [2,7,11,13] target = 9 res = Solution().twoSum(nums,target) print(res) #解法三:暴力法,两层循环遍历,执行效率那自然是不用说了,差的不能再差了,差点要超出时间限制了。 class Solution: def twoSum(self,nums,target): n = len(nums) for x in range(n): for j in range(x+1,n): if nums[j] == target - nums[x]: return x,j nums = [2,7,11,13] target = 9 res = Solution().twoSum(nums,target) print(res)