TwoSum--python

1.TowSum

问题描述：
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.

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

解题过程：
1.初学算法遇到这问题第一反应是做两两相加，循环遍历，时间复杂度O(n^2),这个方法效率很低代码如下：

def towsum(self,num,target):  for i in range(len(num) - 1):      target = target - num[i]      for j  in range(i + 1,len(num)):          if num[j] == target:              return [i,j]

2.优化实现，提高效率，参考其他的博客，有两种优化实现方法，下面一一列出：

1. 排序数组，头索引和尾索引加和，向中间移动索引，直到加和等于目标数值。复杂度为排序的复杂度O(nlog(n))加上比较的复杂度O(n)

   def twosum(self,num,target):    nums = sorted(num)    a = 0    b = len(nums) - 1    index = []    while a < b:        if nums[a] + nums[b] < target:            a = a + 1        elif nums[a] + nums[b] > target:            b = b - 1        elif nums[a] + nums[b] == target:            print(nums[a],nums[b])            for i in range(len(num)):                if nums[a] == num[i]:                    index.append(i)                    break            for j in range(len(num) - 1,-1,-1):                if nums[b] == num[j]:                    index.append(j)                    break            index = sorted(index)            break    return [index[0],index[1]]

2. 用python的字典，也就是哈希，在字典里面加入数组里面的数字，并查找数字，如果查找到就是匹配上了。这个方法我很难想到，对字典的操作，很巧妙，代码简洁，时间复杂度为O(n)代码如下：

   dict = {}for i in range(len(num)):   x = num[i]   if target - x in dict：      return (dict[target - x], i)   dict[x] = i