文本左右对齐

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                  "
]

    来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/text-justification 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution 1

模拟堆栈

class Solution:
    def fullJustify(self, words: [str], maxWidth: int) -> [str]:
        result = []
        while True:
            wordNum = 0
            widthRemain = maxWidth
            while (wordNum < len(words) and widthRemain-wordNum >= len(words[wordNum])):
                widthRemain -= len(words[wordNum])
                wordNum += 1
            if (wordNum < len(words)):
                # normal line
                line = words.pop(0)
                if wordNum == 1:
                    line += ' ' * widthRemain
                    result.append(line)
                else:
                    spaceWidth = int(widthRemain / (wordNum-1))
                    extraSpace = widthRemain % (wordNum-1)
                    for i in range(wordNum-1):
                        if i < extraSpace:
                            line += ' '
                        line += ' ' * spaceWidth
                        line += words.pop(0)
                    result.append(line)
            else:  # last line
                line = words.pop(0)
                for _ in range(wordNum-1):
                    line += ' '
                    line += words.pop(0)
                line += ' ' * (widthRemain - wordNum + 1)
                result.append(line)
                break
        return result