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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:
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
题解
class Solution {public List<String> fullJustify(String[] words, int maxWidth) {List<String> ans = new ArrayList<String>();int right = 0, n = words.length;while (true) {int left = right; // 当前行的第一个单词在 words 的位置int sumLen = 0; // 统计这一行单词长度之和// 循环确定当前行可以放多少单词，注意单词之间应至少有一个空格while (right < n && sumLen + words[right].length() + right - left <= maxWidth) {sumLen += words[right++].length();}// 当前行是最后一行：单词左对齐，且单词之间应只有一个空格，在行末填充剩余空格if (right == n) {StringBuffer sb = join(words, left, n, " ");sb.append(blank(maxWidth - sb.length()));ans.add(sb.toString());return ans;}int numWords = right - left;int numSpaces = maxWidth - sumLen;// 当前行只有一个单词：该单词左对齐，在行末填充剩余空格if (numWords == 1) {StringBuffer sb = new StringBuffer(words[left]);sb.append(blank(numSpaces));ans.add(sb.toString());continue;}// 当前行不只一个单词int avgSpaces = numSpaces / (numWords - 1);int extraSpaces = numSpaces % (numWords - 1);StringBuffer sb = new StringBuffer();sb.append(join(words, left, left + extraSpaces + 1, blank(avgSpaces + 1))); // 拼接额外加一个空格的单词sb.append(blank(avgSpaces));sb.append(join(words, left + extraSpaces + 1, right, blank(avgSpaces))); // 拼接其余单词ans.add(sb.toString());}}// blank 返回长度为 n 的由空格组成的字符串public String blank(int n) {StringBuffer sb = new StringBuffer();for (int i = 0; i < n; ++i) {sb.append(' ');}return sb.toString();}// join 返回用 sep 拼接 [left, right) 范围内的 words 组成的字符串public StringBuffer join(String[] words, int left, int right, String sep) {StringBuffer sb = new StringBuffer(words[left]);for (int i = left + 1; i < right; ++i) {sb.append(sep);sb.append(words[i]);}return sb;}
}来自力扣官方题解