1. Two Sum

题目：

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].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

代码：

LeetCode第一题，今天回国偷来做一下，还是蛮有意思的。

乍一看这个题，当然遍历两遍，稳稳找出结果：

 """遍历两边找出来，时间复杂度O(N^2)"""
    def twoSum(self, nums, target):
        for i in nums:
            for j in range(nums.index(i)+1,len(nums)):
                if(i + nums[j] == target):
                    return [nums.index(i),j]

结果运行的好慢哦：

还好用的是python，网上查了下，java这样写全都报超时错误，时间复杂度太高。

于是，遍历一遍，在剩余列表中找到第二个元素即可：

 """遍历一遍找出来，时间复杂度O(N)"""
    def twoSum2(self, nums, target):
            for i in nums:
                try:
                    #找出第二个元素是否存在，存在返回位置，从当前元素下一位开始找，避免重复
                    j = nums[nums.index(i)+1:].index(target - i)+nums.index(i)+1
                    return [nums.index(i),j]                      
                except:
                    continue

明显快多啦:)

网上有查了下，居然没有想到python的字典，发现用列表写好蠢啊，哎~~~还是要多练习！

边找边存入字典：

"""用字典实现，时间复杂度O(N)"""
    def twoSum3(self, nums, target):
        dict = {}
        for i in range(len(nums)):
            x = nums[i]
            #判断第二个元素的值是否存在在字典中
            if target-x in dict:
                return (dict[target-x], i)
            #将元素按照{key：index}方式存入字典
            dict[x] = i       

速度也明显提升：