12. Integer to Roman
题目描述
罗马数字由七个不同的符号表示:I,V,X,L,C,D和M.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
输入范围是1-3999
例子
2被写成II(2个1相加);12被写成XII(X+II);27被写成XXVII(XX+V+II)。
从大到小,从左向右写。但是,4并不是IIII,而是IV [5-1=4];同理9被写成[IX]。共有6个数字用减法(4/9/40/90/400/900)。
思想
模拟 - 找出数字中的最大的可转换成roman的数字
解法1 (不是最优,改进见解法2)
依次找出数字中包含的最大的可转化为罗马数字的数字。
class Solution(object):
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
nums = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
romans = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
ss = ''
for i in range(len(nums)):
while num >= nums[i]:
num -= nums[i]
ss += romans[i]
return ss
解法2
改进解法1(为了避免很多次的减法运算)
class Solution(object):
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
nums = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
romans = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
ss = ''
for i in range(len(nums)):
if num >= nums[i]:
ss += num // nums[i] * romans[i]
num %= nums[i]
return ss
解法3(最耗时)
刚开始的解法 - 采用二分找到最后一个小于等于num的数
缺点 - 改善的是常数时间复杂度,很冗余。
class Solution(object):
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
mapping = {1:'I', 4:'IV', 5:'V', 9:'IX', 10:'X', 40:'XL', 50:'L', 90:'XC',100:'C', 400:'CD', 500:'D', 900:'CM', 1000:'M'}
lis = [1,4,5,9,10,40,50,90,100,400,500,900,1000]
ss = ''
while num:
pos = self.find1stNum(num, lis)
ss += num // pos * mapping[pos]
num %= pos
return ss
def find1stNum(self, num, lis):
# 找到最后一个小于等于num的数
l = 0
r = len(lis) - 1
while l <= r:
mid = (l + r) >> 1
if lis[mid] <= num:
if mid < len(lis)-1 and lis[mid+1] <= num:
l = mid + 1
else:
return lis[mid]
else:
r = mid - 1
68. Text Justification
题目描述
输入 - 单词数组和宽度maxWidth
要求 -
- 格式化文本,是每行有maxWidth宽度个字符,并完全左右对齐;
- 贪心地分配你的单词,使每行包含尽可能多的单词;必要时填充额外的空格,使宽度达到maxWidth;
- 单词间的空格尽可能均匀分布;如果没法满足均匀分布,左边比右边包含更多的空格;
- 最后一行左对齐,且相邻单词间没有额外的空格。
例子
Example 1:
Input:
words = [“This”, “is”, “an”, “example”, “of”, “text”, “justification.”]
maxWidth = 16
Output:
[
“This is an”,
“example of text”,
"justification. "
]
Example 2:
Input:
words = [“What”,“must”,“be”,“acknowledgment”,“shall”,“be”]
maxWidth = 16
Output:
[
“What must be”,
"acknowledgment ",
"shall be "
]
Explanation: Note that the last line is "shall be " instead of “shall be”,
because the last line must be left-justified instead of fully-justified.
Note that the second line is also left-justified becase it contains only one word.
Example 3:
Input:
words = [“Science”,“is”,“what”,“we”,“understand”,“well”,“enough”,“to”,“explain”,
“to”,“a”,“computer.”,“Art”,“is”,“everything”,“else”,“we”,“do”]
maxWidth = 20
Output:
[
“Science is what we”,
“understand well”,
“enough to explain to”,
“a computer. Art is”,
“everything else we”,
"do "
]
思想
法1 - 首先存储每行包含的单词(单词间有空格间隔,每行包含尽可能多的单词),然后调用函数调整行(空格尽可能均匀分布),最后拼接。
法2 - 改善
- 在循环的每一步,我们都要执行row.append(word)和width -= len(word);
- 当row满了,即width - len(row)[单词间至少用一个空格] - len(word) < 0时,不能添加word;
- 调整行(空格尽可能均匀分布)时,采用row[i%(len(row)-1 or 1)] += ’ ’
- 对最后一行左对齐,用
解法1(不是最简)
class Solution(object):
def fullJustify(self, words, maxWidth):
"""
:type words: List[str]
:type maxWidth: int
:rtype: List[str]
"""
wordLen = list(map(lambda x:len(x), words))
res = []
row = []
i = 0
width = maxWidth
while i < len(words):
if width > wordLen[i]:
width -= (wordLen[i] + 1)
row.append(words[i] + ' ') # 单词间空格间隔
elif width == wordLen[i]:
width -= wordLen[i]
row.append(words[i])
else:
if row[-1][-1] == ' ': # 去掉最后一个单词后的空格
row[-1] = row[-1][:-1]
res.append(self.justifyRow(row, maxWidth))
row = []
width = maxWidth
i -= 1
i += 1
if row:
row = ''.join(row)
res.append(row + ' ' * (maxWidth - len(row)))
return res
def justifyRow(self, row, maxWidth):
if len(row) == 1:
return ''.join(row) + ' ' * (maxWidth - len(row[0]))
# 均匀填充空格
space = maxWidth-len(''.join(row))
m = space // (len(row)-1) # 两两单词间多填充m个空格
n = space % (len(row)-1) # 在前n个空格上, 各添加1个空格
for i in range(n):
row[i] += ' '
return ''.join([word + m * ' ' for word in row[:-1]]) + row[-1]
解法2
简化代码,采用row[i%(len(row)-1 or 1)]和 [’ '.join(row).ljust(maxWidth)]的技巧
class Solution(object):
def fullJustify(self, words, maxWidth):
"""
:type words: List[str]
:type maxWidth: int
:rtype: List[str]
"""
res = []
row = []
width = maxWidth
for word in words:
if width - len(row) - len(word) < 0: # 不能添加word
for i in range(width):
row[i%(len(row)-1 or 1)] += ' '
res.append(''.join(row))
width = maxWidth
row = []
row.append(word)
width -= len(word)
return res + [' '.join(row).ljust(maxWidth)]