文本左右对齐

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algorithms	Hard (52.22%)	334	-

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给定一个单词数组 words 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。

注意:

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:

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输入: words = ["This", "is", "an", "example", "of", "text", "justification."], maxWidth = 16
输出:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

示例 2:

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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                  "
]

提示:

1 <= words.length <= 300
1 <= words[i].length <= 20
words[i] 由小写英文字母和符号组成
1 <= maxWidth <= 100
words[i].length <= maxWidth

Discussion | Solution

解法

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struct Solution;

// @lc code=start
impl Solution {
    /// ## 解题思路
    /// -
    pub fn full_justify(words: Vec<String>, max_width: i32) -> Vec<String> {
        let mut res = vec![];
        let mut line_words: Vec<String> = vec![];
        let mut output_lines = vec![];
        for word in words {
            let words_len = line_words.iter().fold(0, |mut len, w| {
                len += w.len();
                len
            });
            if words_len + line_words.len() + word.len() <= max_width as usize {
                line_words.push(word);
            } else {
                output_lines.push(line_words.clone());

                res.push(format_line_words(&mut line_words, max_width));

                line_words.clear();
                line_words.push(word);
            }
        }
        if line_words.len() > 0 {
            let word_len = line_words.iter().fold(0, |mut len, word| {
                len += word.len();
                len
            });
            let last_line = line_words.join(&" ");
            let remain_space = (max_width as usize) - last_line.len();
            if remain_space > 0 {
                res.push(format!("{}{}", last_line, " ".repeat(remain_space)));
            } else {
                res.push(last_line);
            }
        }

        fn format_line_words(line_words: &mut [String], max_width: i32) -> String {
            let word_count = line_words.len();
            if word_count == 0 {
                return String::new();
            }
            let word_len = line_words.iter().fold(0, |mut len, word| {
                len += word.len();
                len
            });
            let need_spaces = (max_width as usize) - word_len;
            if word_count == 1 {
                return format!("{}{}", line_words[0], &" ".repeat(need_spaces));
            } else {
                let mut i = 0;
                while i < need_spaces {
                    line_words[i % (word_count - 1)].push_str(" ");
                    i += 1;
                }
                return format!("{}", line_words.join(&""));
            }
        }

        res
    }
}
// @lc code=end

文本左右对齐

文本左右对齐

解法

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