[LeetCode] Text Justification (put words in lines with same length)

解答从这里来

比较麻烦的字符串细节实现题。需要解决以下几个问题：

1. 首先要能判断多少个word组成一行：
   这里统计读入的所有words的总长curLen，并需要计算空格的长度。假如已经读入words[0:i]。当curLen + i <=L 且加curLen + 1 + word[i+1].size() > L时，一行结束。

2. 知道一行的所有n个words，以及总长curLen之后要决定空格分配：
   平均空格数：k = (L - curLen) / (n-1)
   前m组每组有空格数k+1：m = (L - curLen) % (n-1)

例子：L = 21，curLen = 14，n = 4
k = (21 - 14) / (4-1) = 2
m = (21 - 14) % (4-1) = 1
A—B–C–D

1. 特殊情况：
   (a) 最后一行：当读入到第i = words.size()-1 个word时为最后一行。该行k = 1，m = 0
   (b) 一行只有一个word：此时n-1 = 0，计算(L - curLen)/(n-1)会出错。该行k = L-curLen, m = 0
   (c) 当word[i].size() == L时.

public ArrayList<String> fullJustify(String[] words, int L) {
    ArrayList<String> res = new ArrayList<String>();
    if(words==null || words.length==0)
        return res;
    //count is the current length of total words length of current round(line)
    int count = 0;
    // last the last word's index from last round (line)
    //well.. it is a actually the first one in this round
    int last = 0;
    for(int i=0;i<words.length;i++)
    {
        //here it is the whole length if adding current word at index i
        //i-last is the # of space, need to ensure each empty spot has at least one space
        if(count+words[i].length()+(i-last)>L)
        {
            int spaceNum = 0;
            int extraNum = 0;
            //if there is more than one word (deal with the special case later)
            if(i-last-1>0)
            {
                //every slot has spaceNum of space
                spaceNum = (L-count)/(i-last-1);
                //the first extraNum of slot has 1 extra space
                extraNum = (L-count)%(i-last-1);
            }
            StringBuilder str = new StringBuilder();
            for(int j=last;j<i;j++)
            {
                str.append(words[j]);
                if(j<i-1)
                {
                    for(int k=0;k<spaceNum;k++)
                    {
                        str.append(" ");
                    }
                    //append an extra space
                    if(extraNum>0)
                    {
                        str.append(" ");
                    }
                    extraNum--;
                }
            }
            //this is only for the special case where there is only one word in the line
            for(int j=str.length();j<L;j++)
            {
                str.append(" ");
            }       
            res.add(str.toString());
            count=0;
            last=i;
        }
        //no matter we reach the end (count is reset to 0) or not (keep increment i), we need to add the length to count.
        count += words[i].length();
    }

    //deal with the last row
    StringBuilder str = new StringBuilder();
    //will only run if there is more word to put in. 
    //It will reach here only because the length "count+words[i].length()+(i-last)" didn't reach the threshold L
    for(int i=last;i<words.length;i++)
    {
        str.append(words[i]);
        if(str.length()<L)
            str.append(" ");
    }
    for(int i=str.length();i<L;i++)
    {
        str.append(" ");
    }
    res.add(str.toString());
    return res;
}