leetcode50_Pow(x, n)

一.问题描述

Implement pow(x, n).

实现指数乘法。

二.代码编写

首先想到的其实就是把n不断拆分成n/2，但是想歪了，可能沉浸在大数乘法那个题里，然后发现其实小数乘大数比两个相等的数运算复杂度低一点，所以就否定了这个想法。但看了tags是二分的思想，后来一想其实重点不在于每次运算的复杂度，而在于二分能将运算的次数由O(N)降低到O(logN)。

所以其实这题很简单啦。重点是不要想偏了！

'''
@ author: wttttt at 2016.11.29
@ problem description see: https://leetcode.com/problems/anagrams/ @ solution explanation see: http://blog.csdn.net/u014265088/article/ @ github:https://github.com/wttttt-wang/leetcode
@ hash table, use python dictionary
@ its better to use sorted(i) as the key
'''
class Solution(object):
def groupAnagrams(self, strs):
"""
:type strs: List[str]
:rtype: List[List[str]]
"""
d,ans= {},[]
for i in strs:
sortstr = ''.join(sorted(i)) # use sorted(i) as the key in the dictionary
if sortstr in d:
d[sortstr] += [i]
else:
d[sortstr] = [i]
#print(d)
for i in d:
tmp = d[i];tmp.sort()
ans += [tmp]
return ans

so = Solution()
strs = ["eat", "tea", "tan", "ate", "nat", "bat"]
print so.groupAnagrams(strs)