[LeetCode]68 文本左右对齐

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:

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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                  "
]

代码

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class Solution:
    def fullJustify(self, words, maxWidth):
        """
        :type words: List[str]
        :type maxWidth: int
        :rtype: List[str]
        """
        if not words:
            return list()
        
        res, row = list(), list()
        words_len, count = len(words), 0
        i = 0
        while i < words_len:
            if len(words[i]) + count <= maxWidth:
                count += len(words[i]) + 1
                row.append(words[i])
                i += 1
                continue
            else:
                if len(row) == 1:
                    res.append(str(row[0])+' '*(maxWidth - len(row[0])))
                else:
                    extraSpace = maxWidth - count + len(row)
                    num_space = extraSpace//(len(row) - 1)
                    pre_space = extraSpace%(len(row) - 1)
                    tmp = ''
                    for word in row[:-1]:
                        if pre_space > 0:
                            tmp += str(word) + ' '*(num_space + 1)
                            pre_space -= 1
                        else:
                            tmp += str(word) + ' '*(num_space + 0)
                            
                    tmp += str(row[-1])
                    res.append(tmp)
                    
                row.clear()
                count = 0
                
        tmp = ' '.join(row)  
        res.append(tmp + (maxWidth - len(tmp))*' ')
        return res