本题考查细节处理和根据要求写出程序的能力. 思路就是顺序扫描词, 一行一行将词放入. 对于每一行, 先求出这行最多能放下多少个单词, 再根据是不是最后一行决定是left justification还是平均分布空格.
注意过程中有多处需要巧妙构思的写法. 比如怎么平均分布空格. 具体看代码
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class { public List<String> fullJustify(String[] words, int maxWidth) { List<String> res = new ArrayList<>(); int wordIdx = 0; while (wordIdx < words.length) { StringBuilder sb = new StringBuilder(); int count = words[wordIdx].length(); int lastWordIdx = wordIdx + 1; while (lastWordIdx < words.length) { if (count + 1 + words[lastWordIdx].length() > maxWidth) break; count += 1 + words[lastWordIdx].length(); lastWordIdx++; } sb.append(words[wordIdx]); int numOfWordsLeft = lastWordIdx - wordIdx - 1; if (lastWordIdx == words.length || numOfWordsLeft == 0) { for (int i = wordIdx + 1; i < lastWordIdx; i++) { sb.append(' '); sb.append(words[i]); } for (int i = sb.length(); i < maxWidth; i++) { sb.append(' '); } } else { int spaces = (maxWidth - count) / numOfWordsLeft; int extraSpace = (maxWidth - count) % numOfWordsLeft; for (int i = wordIdx + 1; i < lastWordIdx; i++) { for (int j = 0; j < spaces; j++) { sb.append(' '); } if (extraSpace > 0) { sb.append(' '); extraSpace--; } sb.append(' '); sb.append(words[i]); } } wordIdx = lastWordIdx; res.add(sb.toString()); } return res; } }
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