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 the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
首先考虑到时间复杂度O(N^2)的循环遍历，leetcode不能忍。
第二种方案复杂度为O(N)，遍历数组时，以数值为key，索引为value建立字典，然后每次target值减当前值从字典中获取index，index小于i则说明在字典中存在。
代码O(N^2)
class Solution1: # @param {integer[]} nums # @param {integer} target # @return {integer[]} def twoSum(self, nums, target): rlt = [] if not nums: return rlt for i in range(len(nums)): for j in range(len(nums[i+1:])): if nums[i] + nums[i+j] == target: rlt.append(i) rlt.append(i+j) return rlt
class Solution: # @param {integer[]} nums # @param {integer} target # @return {integer[]} def twoSum(self, nums, target): rlt = [] if not nums: return rlt num_dict = {} for i in range(len(nums)): if nums[i] not in num_dict: num_dict[nums[i]] = i print nums[i],i j = num_dict.get(target-nums[i], -1) if j < i and j > -1: rlt.append(j+1) rlt.append(i+1) return rlt return rlt
原文：http://www.cnblogs.com/ottll/p/4493429.html