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:
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                  "
]
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解题思路: 贪心

1. 使用贪心的思路, 尽可能的让一行放下更多的单词
2. 当这一行放不下下一个单词时, 就重新开一行. 这样就可以得到每一行有哪些单词
3. 根据每一行的单词, 填充空格
4. 最后一行做单独处理

示例代码

def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
    res, cur, l = [], [], 0
    # 根据贪心, 计算每一行有哪些单词
    for item in words:
        if l + len(item) <= maxWidth:
            l += (len(item) + 1)
            cur.append(item)
        else:
            res.append(cur)
            l = len(item) + 1
            cur = [item]
    res.append(cur)

    # 根据每一行的单词进行空格填充
    ans = []
    for items in res[:-1]:
        count = len(items)
        if count == 1:
            ans.append(items[0] + " " * (maxWidth - len(items[0])))
        elif count == 2:
            ans.append(items[0] + " " * (maxWidth - len(items[0]) - len(items[1])) + items[1])
        else:
            sumC = sum(len(i) for i in items)
            m = (maxWidth - sumC) // (len(items) - 1)
            n = (maxWidth-sumC) - m*(len(items) - 1)
            a = items[0]
            i = 1
            for w in items[1:]:
                if n > 0:
                    a = a + " " * (m+1) + w
                else:
                    a = a + " " * m + w
                n -= 1

            ans.append(a)

    # 处理最后一行
    items = res[-1]
    a = items[0]
    for w in items[1:]:
        a += (" " + w)
    a = a + (" " * (maxWidth - len(a)))
    ans.append(a)

    return ans

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