LeetCode-1- 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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

 

简言之，查找给定数组内两个数，使得两数和为给定的target

 

思路：

　　遍历给定数组，用一个哈希表存放已经遍历过的值，key是nums中的值，value是其位置索引。

　　对于当前遍历的nums[i]，查找哈希表中是否存在target-nums[i]，若存在，则返回二者索引，否则将(key=nums[i],value=i)放入哈希表中，继续遍历

 

时间复杂度：O(n)

 

python代码如下，已AC:

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        dictMap = {}        for index, value in enumerate(nums):            if target - value in dictMap:                return [dictMap[target - value] , index ]            dictMap[value] = index

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