leetcode 68 Text Justification

Lc68 Text Justification

我们首先要确定当前这一行能放下words[i]~words[?]，

然后是如何分配空格，

如果是最后一行或是这一行只能放下一个单词，那么空格应该全部在右侧

否则，应该平均分配，余下的依次从左到右分配

class Solution {
    public List<String> fullJustify(String[] words, int maxWidth) {
        int index = 0;
        List<String> res = new ArrayList<>();
        while(index < words.length){
            int last = index + 1;//index~last-1为这一行能放下的单词
            int count = words[index].length();
            while(last < words.length){
                if(count + 1 + words[last].length() > maxWidth)
                    break;
                count += words[last].length() + 1;
                last++;
            }
            
            StringBuilder sb = new StringBuilder();
            sb.append(words[index]);//先将index放入这一行
            int lineWordsCnt = last - index - 1;//剩下未放单词的数量
            if(lineWordsCnt == 0 || last == words.length){
                for(int i=index+1; i<last; i++){
                    sb.append(" ").append(words[i]);
                }
                for(int i=count; i<maxWidth; i++){
                    sb.append(" ");
                }
            }else{
                int spaces = (maxWidth - count) / lineWordsCnt; //每个空挡都应该放的空格
                int plusSpaces = (maxWidth - count) % lineWordsCnt; //未除尽，余下的空格，依次从左到右填上
                
                for(int i=index+1; i<last; i++){
                    for(int j=0; j<spaces; j++){
                        sb.append(" ");
                    }
                    if(plusSpaces > 0){
                        sb.append(" ");
                        plusSpaces--;
                    }
                    sb.append(" ");
                    sb.append(words[i]);
                }
            }
            res.add(sb.toString());
            index = last;
        }
        return res;
    }
}