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20鱼币
今天的题目:
给定一个单词数组和一个长度 maxWidth,重新排版单词,使其成为每行恰好有 maxWidth 个字符,且左右两端对齐的文本。
你应该使用 “贪心算法” 来放置给定的单词;也就是说,尽可能多地往每行中放置单词。必要时可用空格 ' ' 填充,使得每行恰好有 maxWidth 个字符。
要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配,则左侧放置的空格数要多于右侧的空格数。
文本的最后一行应为左对齐,且单词之间不插入额外的空格。
说明:
1. 单词是指由非空格字符组成的字符序列。
2. 每个单词的长度大于 0,小于等于 maxWidth。
3. 输入单词数组 words 至少包含一个单词。
示例 1:
输入:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
输出:
[
"This is an",
"example of text",
"justification. "
] 示例 2:
输入:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
输出:
[
"What must be",
"acknowledgment ",
"shall be "
]
解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
因为最后一行应为左对齐,而不是左右两端对齐。
第二行同样为左对齐,这是因为这行只包含一个单词。 示例 3:
输入:
words = ["Science","is","what","we","understand","well","enough","to","explain",
"to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
输出:
[
"Science is what we",
"understand well",
"enough to explain to",
"a computer. Art is",
"everything else we",
"do "
]
 欢迎大家一起答题! 
- def f253(words, maxWidth):
- def adjustSpace(lst):
- string = ""
- num_lst, len_lst = len(lst), len(''.join(lst))
- if num_lst == 1:
- return lst[0] + ' ' * (maxWidth - len_lst)
- every_space = (maxWidth - len_lst) // (num_lst - 1)
- surplus = (maxWidth - len_lst) % (num_lst - 1)
- for i in range(num_lst - 1):
- string += lst[i] + ' ' * every_space
- if surplus:
- string += ' '
- surplus -= 1
- return string + lst[-1]
- def adjustLastLine(lst):
- if len(lst) == 1:
- return lst[0] + ' ' * (maxWidth - len(lst[0]))
- string = ' '.join(lst)
- return string + ' ' * (maxWidth - len(string))
- res = []
- list_tmp, len_list_tmp = [], 0
- for word in words:
- len_word = len(word)
- if len_list_tmp + len_word < maxWidth:
- list_tmp.append(word)
- len_list_tmp += len_word + 1
- elif len_list_tmp + len_word == maxWidth:
- list_tmp.append(word)
- res.append(' '.join(list_tmp))
- list_tmp, len_list_tmp = [], 0
- else:
- res.append(adjustSpace(list_tmp))
- list_tmp, len_list_tmp = [word], len_word + 1
- if len_list_tmp:
- res.append(adjustLastLine(list_tmp))
- return res
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