68. 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                  "
]

解：hard,题目思路应该不难，先判断一行最多能存放多少个单词，1个单词0个空格，2个单词1+个空格，3个单词2+个空格…

class Solution {
    public List<String> fullJustify(String[] words, int L) {
        List<String> lines = new ArrayList<String>();

        int index = 0;
        while (index < words.length) {
            int count = words[index].length();
            int last = index + 1;
            while (last < words.length) {
                if (words[last].length() + count + 1 > L) {
                    break;
                }
                count += words[last].length() + 1;
                last++;
            }

            StringBuilder builder = new StringBuilder();
            int diff = last - index - 1;
            // if last line or number of words in the line is 1, left-justified
            if (last == words.length || diff == 0) {
                for (int i = index; i < last; i++) {
                    builder.append(words[i] + " ");
                }
                builder.deleteCharAt(builder.length() - 1);
                for (int i = builder.length(); i < L; i++) {
                    builder.append(" ");
                }
            } else {
                // middle justified
                int spaces = (L - count) / diff;
                int r = (L - count) % diff;
                for (int i = index; i < last; i++) {
                    builder.append(words[i]);
                    if (i < last - 1) {
                        for (int j = 0; j <= (spaces + ((i - index) < r ? 1 : 0)); j++) {
                            builder.append(" ");
                        }
                    }
                }
            }
            lines.add(builder.toString());
            index = last;
        }
        return lines;
    }
}

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