文本左右对齐

题目

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

题解

这题太恶心了，不想写了，操

public List<String> fullJustify(String[] words, int maxWidth) {
    int head = 0;
    List<String> result = new ArrayList<>();
    while (head < words.length) {
        int[] nHaL = findNH(head, maxWidth, words);
        if (nHaL[0] == words.length) {
            StringBuilder sb = new StringBuilder();
            for (int i = head; i < words.length; i++) {
                sb.append(words[i]).append(" ");
            }
            if(sb.length() >= maxWidth){
                sb.deleteCharAt(sb.length() - 1);
            }
            while(sb.length() < maxWidth){
                sb.append(" ");
            }
            result.add(sb.toString());
        } else {
            StringBuilder sb = hanlder(head, nHaL, words, maxWidth);
            result.add(sb.deleteCharAt(sb.length() - 1).toString());
        }
        head = nHaL[0];
    }
    return result;
}

private int[] findNH(int head, int maxWidth, String[] words) {
    int curlength = 0;
    int i = head;
    int[] result = new int[2]; // 0: next head; 1: curlength;
    while (i < words.length) {
        String nW = words[i];
        curlength += nW.length();
        if (curlength > maxWidth) {
            curlength -= nW.length();
            result[0] = i;
            result[1] = curlength - 1;
            return result;
        }
        curlength += 1; // for the space
        i++;
    }
    result[0] = i;
    result[1] = curlength - 1;
    return result;
}

private StringBuilder hanlder(int oh, int[] nh, String[] words, int maxWidth) {
    int head = oh;
    int tail = nh[0];
    StringBuilder sb = new StringBuilder();
    double tspace = (double) (maxWidth - nh[1]);
    double gaps = (double) (tail - head - 1);
    if(gaps == 0){
        sb.append(words[head]);
        while(sb.length() < maxWidth){
            sb.append(" ");
        }
        return sb.append(" ");
    }
    while (head < tail) {
        sb.append(words[head]).append(" ");
        int extra = (int) Math.ceil(tspace / gaps);
        tspace -= extra;
        while (extra > 0) {
            sb.append(" ");
            extra--;
        }
        gaps--;
        head++;
    }
    return sb;
}