I created the OASIS because I never felt at home in the real world. I didn’t know how to connect with the people there. I was afraid, for all of my life, right up until I knew it was ending. That was when I realized, as terrifying and painful as reality can be, it’s also the only place where you can find true happiness. Because reality is real.
words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.Input:["abcw","baz","foo","bar","xtfn","abcdef"]Output:16 Explanation:The two words can be"abcw", "xtfn".
Example 2:
Input:["a","ab","abc","d","cd","bcd","abcd"]Output:4 Explanation:The two words can be"ab", "cd".
Example 3:
Input:["a","aa","aaa","aaaa"]Output:0 Explanation:No such pair of words.
Input: [4,3,2,7,8,2,3,1] Output: [5,6]
刚一看到这道题是Medium难度的时候，我觉得Leetcode高估了这道题。但是后面O(1)空间和O(n)时间发现还是没有那么容易的。
seen = list()
result = list()
for item in nums:
if item in seen:
result += [item]
seen += [item]
else:
seen += [item]
return result
这种方法按理说是可以过的 (虽然不符合题目对空间复杂度的要求)，但是在Leetcode平台上，我发现这种方法会超时(TLE)
res = []
for x in nums:
if nums[abs(x) - 1] < 0:
res.append(abs(x))
else:
nums[abs(x) - 1] *= -1
return res
想了十分钟之后，我发现我貌似想不出什么好方法了，于是看了一下Discuss里大佬的解法。
这个解法真的是Awesome！在保证了O(n)时间的前提下，完美利用之前的空间做了一次Hash。只能说想出这个方法的人是个小天才了！(这里是Disscuss的帖子地址)
我想让你们见识一下什么是真正的勇敢
勇敢不是一个人拿着枪
勇敢是当你还未开始就已经知道自己会输
可你依然坚持去做
而且无论如何都要把它坚持到底
你很少能赢
但有时候也会
It is an immemorial game which I really hope you can enjoy.
#coding: utf-8
import time
import numpy as np
import subprocess
class LifeSpace:
def __init__(self, name, israndom, size_x, size_y):
self.name = name
self.israndom = israndom
self.size = (size_x, size_y)
self.iter_times = 0
self.Z = Z = np.random.randint(0, 2, self.size)
def iterate(self):
# Count neighbours
N = (self.Z[0:-2, 0:-2] + self.Z[0:-2, 1:-1] + self.Z[0:-2, 2:] +
self.Z[1:-1, 0:-2] + self.Z[1:-1, 2:] +
self.Z[2: , 0:-2] + self.Z[2: , 1:-1] + self.Z[2: , 2:])
# Apply rules
birth = (N == 3) & (self.Z[1:-1, 1:-1] == 0)
survive = ((N == 2) | (N == 3)) & (self.Z[1:-1, 1:-1] == 1)
self.Z[...] = 0
self.Z[1:-1, 1:-1][birth | survive] = 1
# Iterate update
self.iter_times += 1
def get_Z_status(self):
print("--"*(self.size[0]/2) + str(self.iter_times) + "--"*(self.size[0]/2))
line = ''
for i in self.Z:
for j in i: line += "■" if j == 1 else "□"
print line
line = ''
def process(self, iter_time):
for i in range(iter_time):
self.iterate()
self.get_Z_status()
time.sleep(0.3)
subprocess.call("clear")
if __name__ == "__main__":
newspace = LifeSpace("Elf", True, 15, 30)
newspace.process(1000)
BTW, You can also find it at Github.
这两天鱼神要去面Google，居然破天荒的准备了起来。 昨天晚上他突然给我发了一道Leetcode上的Easy题 —— Climbing Stairs。我心想按照鱼神的水准，这种题怎么可能入得了他的法眼。后来看了看其实这道题还是挺能看出一个人的水平的（突然想起了求素数） 这道题由于比较基础，所以解法相对来说也是比较多的。首先先上题目描述：
You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n will be a positive integer.
因为在学习动态规划的时候，这道题就是用来入门的，所以一拿到这道题，应该很自然的就可以想到动态规划的方法。 状态转移方程为：ways[n] = ways[n – 1] + ways[n – 2] 然后套路的发现这个形式长的居然和斐波那契数列一摸一样（对！没错，这题就是套着斐波那契出的。）所以就产生下面的几种解法：
Method 1: 递归求解
class Solution(object):
def climbStairs(self, n):
if n == 1 or n <= 0:
return 1
return climbStairs(n-1) + climbStairs(n-2)
Method 2: 非递归求解
class Solution(object):
def climbStairs(self, n):
a = b = 1
for x in range(2, n + 1):
a, b = b, a + b
return b
Method 3：通项公式求解
class Solution(object):
def climbStairs(self, n):
sqrt5 = math.sqrt(5)
Phi = (1 + sqrt5) / 2
phi = (1 - sqrt5) / 2
return int((Phi ** (n + 1) - phi ** (n + 1)) / sqrt5)
Method 4: 杨辉三角叠加法
class Solution(object):
def c(self, m, n):
x = 1
for i in xrange(m-n+1, m+1):
x *= i
for i in xrange(2, n+1):
x /= i
return x
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
result = 0
for i in range(0, n/2+1):
result += self.c(n-i, i)
return result
今天因为需要把尘封了好久的电脑拿了出来
手碰到键盘的一瞬间
感觉宇宙星河呼啸而来
你是我未曾拥有无法捕捉的亲昵
我却有你的吻你的魂你的心
载着我飞呀飞呀飞 越过了意义
你是我朝夕相伴触手可及的虚拟
/*
int total = 0;
for (int i = 0; i < doc['goals'].length; ++i)
{
total += doc['goals'][i];
}
return total;
*/
人生不如意之事七八九，苦事
终归还能与人言一二三，幸事
@花花