文本左右对齐

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

class Solution {
    public List<String> fullJustify(String[] words, int maxWidth) {
        /**
         * special:
         * lastLine or only one word in a line
         */
        int max, len = words.length, start, end = 0, spaceNum, wordNum;
        List<String> res = new ArrayList<>();
        while (true) {
            start = end;
            max = 0;
            while (end < len && max + words[end].length()+(end-start) <= maxWidth){
                max += words[end++].length();
            }
            /**
             * lastLine
             */
            if (end == len){
                StringBuffer sb = new StringBuffer(words[start]);
                for (int i = start+1; i < end; i++) {
                    sb.append(' ');
                    sb.append(words[i]);
                }
                int length = sb.length();
                for (int i = 0; i < maxWidth-length; i++) {
                    sb.append(' ');
                }
                res.add(sb.toString());
                return res;
            }
            wordNum = end - start;
            spaceNum = maxWidth - max;
            /**
             * only one word in a line
             */
            if (wordNum == 1) {
                StringBuffer sb = new StringBuffer(words[start]);
                for (int i = 0; i < spaceNum; i++) {
                    sb.append(' ');
                }
                res.add(sb.toString());
                continue;
            }

            /**
             * common situation
             */
            StringBuffer sb = new StringBuffer(words[start]);
            for (int i = start+1; i < end; i++) {
                if (spaceNum % (wordNum-1) == 0) {
                    for (int j = 0; j < spaceNum/ (wordNum-1); j++) {
                        sb.append(' ');
                    }
                }else {
                    spaceNum -= 1;
                    sb.append(' ');
                    for (int j = 0; j < spaceNum/ (wordNum-1); j++) {
                        sb.append(' ');
                    }
                }
                sb.append(words[i]);
            }
            res.add(sb.toString());
        }
    }
}