Leetcode 68. 文本左右对齐

题干

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:

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                  "
]

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

初见

读假题了......看到第一个样例以为要均匀分布......

优化

根据贪心，计算当前行最多能容纳的单词数，并计算需要填充的空格数即可，注意特判最后一行。感觉没有 hard 难度。

class Solution {
    public List<String> fullJustify(String[] words, int maxWidth) {
        List<String> answer = new ArrayList<String>();
        int right = 0;
        while (true) {
            int left = right;
            int lineLength = 0;
            // 当前行能容纳的单词数
            while (right < words.length && lineLength + words[right].length() + right - left <= maxWidth) {
                lineLength += words[right++].length();
            }
            // 最后一行单词间固定只填充一个空格，剩余空格填充在行尾
            if (right == words.length) {
                StringBuffer stringBuffer = join(words, left, words.length, " ");
                stringBuffer.append(blank(maxWidth - stringBuffer.length()));
                answer.add(stringBuffer.toString());
                return answer;
            }

            int wordsNum = right - left;
            int spacesNum = maxWidth - lineLength;
            // 只有一个单词，空格全部填充在行尾
            if (wordsNum == 1) {
                StringBuffer stringBuffer = new StringBuffer(words[left]);
                stringBuffer.append(blank(spacesNum));
                answer.add(stringBuffer.toString());
                continue;
            }
            // 分配相同空格数给每个单词，若有余数，前余数个需要再多一个空格
            int avgSpaces = spacesNum / (wordsNum - 1);
            int extraSpaces = spacesNum % (wordsNum - 1);
            StringBuffer stringBuffer = new StringBuffer();
            stringBuffer.append(join(words, left, left + extraSpaces + 1, blank(avgSpaces + 1)));
            stringBuffer.append(blank(avgSpaces));
            stringBuffer.append(join(words, left + extraSpaces + 1, right, blank(avgSpaces)));
            answer.add(stringBuffer.toString());
        }
    }

    public String blank(int n) {  // 包含n个空格的字符串
        StringBuffer stringBuffer = new StringBuffer();
        for (int i = 0; i < n; ++i) {
            stringBuffer.append(' ');
        }
        return stringBuffer.toString();
    }

    public StringBuffer join(String[] words, int left, int right, String sep) {  // 在单词间填充不同长度的空格串
        StringBuffer stringBuffer = new StringBuffer(words[left]);
        for (int i = left + 1; i < right; ++i) {
            stringBuffer.append(sep);
            stringBuffer.append(words[i]);
        }
        return stringBuffer;
    }
}