1. Unique Morse Code Words
1.1 题目描述
给定一个字符串数组，利用摩斯密码表将字符串加密，找出经加密后，不同摩斯密码的个数
Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation:
经过转换的摩斯密码有：
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."
有两种相异的摩斯密码，分别是："--...--." 和 "--...-."
1.2 思路&解法
简单模拟：
class Solution(object):
def uniqueMorseRepresentations(self, words):
"""
:type words: List[str]
:rtype: int
"""
List = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
result = ""
Set = set()
for word in words:
for c in word:
index = ord(c) - ord('a')
result = result + List[index]
Set.add(result)
result = ""
return len(Set)
标准答案的实现代码如下：
class Solution(object):
def uniqueMorseRepresentations(self, words):
MORSE = [".-","-...","-.-.","-..",".","..-.","--.",
"....","..",".---","-.-",".-..","--","-.",
"---",".--.","--.-",".-.","...","-","..-",
"...-",".--","-..-","-.--","--.."]
seen = {"".join(MORSE[ord(c) - ord('a')] for c in word)
for word in words}
return len(seen)
.join()方法用于将序列中的元素以指定的字符连接生成一个新的字符串。
充分利用了python中迭代器的机制，但是两个答案思想都是一样的。
2. Max Increase to Keep City Skyline
2.1 问题描述
一个二位数组，数组的每一个元素都代表着一座楼房的高度。skyline代表着从四个方向看去，必须和原始的skyline一致。简单来说，我们用不同的角度去看这个二维矩阵，都有一个最大值的数组作为上限值。在保持这两个上限值数组不变的条件下，我们要将各个建筑的高度最大化，并且计算出各个建筑增加高度的和。
例如：
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation:
The grid is:
[ [3, 0, 8, 4],
[2, 4, 5, 7],
[9, 2, 6, 3],
[0, 3, 1, 0] ]
The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]
The grid after increasing the height of buildings without affecting skylines is:
gridNew = [ [8, 4, 8, 7],
[7, 4, 7, 7],
[9, 4, 8, 7],
[3, 3, 3, 3] ]
2.2 思路与解答
初始思路：
初始思路是要先构造一个转置的矩阵，再构造两个skyline数组，最后根据两个skyline数组增加楼的高度。
经过优化后，两个skyline的数组可以同时进行计算。这里利用了zip函数巧妙地将矩阵进行了转置，并且利用enumerate构造了一个可迭代序列。
def maxIncreaseKeepingSkyline(self, grid):
row_maxes = [max(row) for row in grid]
col_maxes = [max(col) for col in zip(*grid)]
return sum(min(row_maxes[r], col_maxes[c]) - val
for r, row in enumerate(grid)
for c, val in enumerate(row))
