Text Justification

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Problem:

Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactlyL characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

For example,
words: ["This", "is", "an", "example", "of", "text", "justification."]
L: 16.

Return the formatted lines as:

[   "This    is    an",   "example  of text",   "justification.  "]

Note: Each word is guaranteed not to exceed L in length.

click to show corner cases.

Corner Cases:

A line other than the last line might contain only one word. What should you do in this case?
In this case, that line should be left-justified.

规则描述了一大堆，我就是按照Word编辑中的两端对齐排版来理解的。关键是对问题中“word”的理解，""算不算一个“word”？从出题人给出的test case来看，显然是把""当成了一个“word”。如此一来，

fullJustify(["is","","a"],6)的输出是["is a"]，还是["is ","a "]。如果是前者，在处理""时直接略过就行了。如果是后者那就把""当一个正常的单词处理。从对齐的效果来看，显然前者更受青睐。

Solution:

public class Solution {
public List<String> fullJustify(String[] words, int L) {
List<String> res = new ArrayList<>();
   if(words==null||words.length==0||L<0)
       return res;
int b=0,e=-1,w=0;
for(int i=0;i<words.length;i++)
{
        if(w==0||w+e-b+words[i].length()+1<=L)
        {
            w += words[i].length();
            e++;
        }
         else
        {
            StringBuilder str = new StringBuilder();
             if(e==b)
            {
                str.append(words[b]);
                int tmp = L - w;
                while(tmp-->0)
                str.append(" ");
            }
            else
            {
                str.append(words[b]);
                int d = (L-w)/(e-b);
                int r = (L-w)%(e-b);
                for(int j=b+1;j<=e;j++)
                {
                    int tmp = d + (r-->0?1:0);
                    while(tmp-->0)
                        str.append(" ");
                    str.append(words[j]);
                }
            }
            w = words[i].length();
            b = i;
            e = i;
            res.add(str.toString());
        }
}
if(w>0||res.size()==0)
{
     StringBuilder str = new StringBuilder();
     str.append(words[b]);
            for(int j=b+1;j<=e;j++)
            {
                 str.append(" ");
                str.append(words[j]);
             }
            int tmp = L - (w + e - b);
            while(tmp-->0)
                str.append(" ");
            res.add(str.toString());
}
return res;
}
}

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