文本左右对齐
- 难度： 困难
- 通过率： 21.9%
- 题目链接：https://leetcode-cn.com/problems/text-justification
题目描述
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:
- 单词是指由非空格字符组成的字符序列。
- 每个单词的长度大于 0，小于等于 maxWidth。
- 输入单词数组
words至少包含一个单词。
示例:
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 " ]
解法：
再也不想做 hard 类型的题了。
class Solution:
def fullJustify(self, words, max_width):
"""
:type words: List[str]
:type maxWidth: int
:rtype: List[str]
"""
rows = []
row_words = []
row_length = 0
row_words_length = 0
for word in words:
word_length = len(word)
if row_length + word_length <= max_width:
row_words_length += word_length
row_length += word_length + 1
row_words.append(word)
else:
rows.append((row_words, row_words_length))
row_words = [word]
row_words_length = word_length
row_length = word_length + 1
rows.append((row_words, row_words_length))
return self.merge(rows, max_width)
def merge(self, rows, max_width):
str_rows = []
for row_count, row in enumerate(rows):
row_words = row[0]
row_words_length = row[1]
words_count = len(row_words)
space_count = max_width - row_words_length
gap_count = words_count - 1
if gap_count == 0:
gap_space_count = 0
remain_space_count = 0
else:
gap_space_count = space_count // gap_count
remain_space_count = space_count % gap_count
if row_count == len(rows) - 1:
gap_space_count = 1
remain_space_count = 0
row_str_list = []
for i, word in enumerate(row_words):
row_str_list.append(word)
if i == len(row_words) - 1:
break
row_str_list.append(' ' * gap_space_count)
if remain_space_count > 0:
row_str_list.append(' ')
remain_space_count -= 1
str_row = ''.join(row_str_list)
# 处理一行只有一个单词或最后一行的情况
if max_width != len(str_row):
str_row += (' ' * (max_width - len(str_row)))
str_rows.append(str_row)
return str_rows