【leetcode】Two Sum

题目简述：

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

解题思想：

首先我很是不甘心的直接暴力了一次，果断TLE。然后想到先排序，然后两个指针前后搜，大了就后面的指针前移，小了就前面的指针后移。不过比较麻烦的是要找出原来的位置，导致代码多了几个丑陋的循环。

#coding=utf-8
class Solution:
    # @return a tuple, (index1, index2)
    def twoSum(self, num, target):
        t = []
        for i in num:
            t.append(i)
        num.sort()
        l = len(num)
        index1 = 0
        index2 = l-1
        while index1 < index2:
            if num[index1] + num[index2] == target:
                break
            elif num[index1] + num[index2] > target:
                index2 -= 1
            elif num[index1] + num[index2] < target:
                index1 += 1
        for i in range(l):
            if num[index1] == t[i]:
                index1 = i
                break
        for i in range(l):        
            if num[index2] == t[i] and i != index1:
                index2 = i
                break
        if index1 >index2:
            return index2+1,index1+1
        return index1+1,index2+1

不过后来看到的标准解法里用了哈希表，We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index。于是就用dict写了下，不过可能是python的dict比较慢还是怎么，速度几乎一样，不过代码倒是短了不少。

class Solution:
    # @return a tuple, (index1, index2)
    def twoSum(self, num, target):
        t = {}
        l = len(num)
        for i in range(l):
            t[num[i]] = i
        for i in range(l):
            if t.has_key(target-num[i]):
                if i != t[target-num[i]]:
                    return i+1 , t[target-num[i]]+1