Text Justification
这一题主要考察的还是对边界条件的控制,并没有太高深的算法,但是编写的难度不低,而且想一次通过非常难。
我采用的办法:
首先,筛选出一行中应该包括的单词,这里要考虑空格还要占用掉的长度,也就是除了当前的总长度之外,还需要记录下纯单词的长度,这样方便之后生成相应的空格。
然后,当当前记录的长度大于等于额定长度之后,就需要对当前内容进行格式化。单词的索引很容易算出,难点在于空格,因为空格需要遵循以下两个规则:
1. 每两个单词之间必须要有一个空格
2. 单词实现左右对齐之后,空格应该尽量平均,并且有需要的话越往左空格应该越多
基于这两条原则,我的办法是先计算平均空格长度,那么所有的空格要么是等于平均空格长度,要么是等于平均空格长度加1,在拼接空格的时候,计算当前剩余的空格总数以每个位置填入平均空格数是否可以填满,以此来判断具体当前位置需要填入的空格数。
实现代码:
java
public class Solution { public List<String> fullJustify(String[] words, int L) { int index = 0, last = 0, count = 0, wordsLenCount = 0, wordsCount = 0; List<String> result = new ArrayList<String>(); if (L == 0) { result.add(words[0]); return result; } while (index < words.length) { if (count + words[index].length() >= L) { int spaceCount; int aveSlot = 0; if (count + words[index].length() > L) { spaceCount = L - wordsLenCount; if (wordsCount > 1) aveSlot = spaceCount / (wordsCount - 1); index--; } else { wordsLenCount += words[index].length(); wordsCount++; spaceCount = L - wordsLenCount; if (wordsCount > 1) aveSlot = spaceCount / (wordsCount - 1); } StringBuffer buffer = new StringBuffer(""); for (int i = index - wordsCount + 1; i <= index; i++) { buffer.append(words[i]); int sCount; if (wordsCount > 1) { if (spaceCount % aveSlot == 0 && spaceCount / aveSlot == wordsCount - 1) { sCount = aveSlot; } else { sCount = aveSlot + 1; } spaceCount -= sCount; wordsCount--; } else { sCount = spaceCount; } for (int j = 0; j < sCount; j++) { buffer.append(' '); } } result.add(buffer.toString()); count = 0; wordsLenCount = 0; wordsCount = 0; last = index + 1; } else { count += words[index].length() + 1; wordsLenCount += words[index].length(); wordsCount++; } index++; } if (last < words.length) { StringBuffer buffer = new StringBuffer(""); for (int i = last; i < words.length; i++) { buffer.append(words[i]); if (words[i].length() > 0) buffer.append(" "); } for (int i = buffer.length(); i < L; i++) { buffer.append(" "); } result.add(buffer.toString()); } return result; } }
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