LeetCode-Text Justification

Text Justification

##题目

####Text Justification

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 exactly L 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.

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.

##解题思路
这道题属于纯粹的字符串操作，要把一串单词安排成多行限定长度的字符串。
主要难点在于空格的安排，首先每个单词之间必须有空格隔开，而当当前行放不下更多的单词并且字符又不能填满长度L时，我们要把空格均匀的填充在单词之间。如果剩余的空格量刚好是间隔倍数那么就均匀分配即可，否则还必须把多的空格依次从左往右放到前面的间隔里面。最后一个细节就是最后一行不需要均匀分配空格，句尾留空就可以，所以要单独处理一下。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public ArrayList<String> fullJustify(String[] words, int L) {
        ArrayList<String> res = new ArrayList<String>();  
	    if(words==null || words.length==0)  
	        return res;  
	    int first=0; //一行中第一个单词的位置
	    int last=0;  //一行中最后一个单词的位置
	    int count=0; //一行中的字符数(不包括空格)
	    for(;last<words.length;last++)
	    {
	    	if((count+words[last].length()+last-first)>L)  //last-first表示单词间隔数,因为单词之间至少一个空格
	    	{
	    		last--; //最后一个单词不满足
	    		int konggenum=L-count;
	    		//判断空格是否可以被间隔均分
	    		int isjunfen=0;
	    		int countkongge=0;
	    		if((last-first)>0)
	    		{
	    			isjunfen=konggenum%(last-first);
	    			countkongge=konggenum/(last-first);
	    		}
	    		StringBuilder row=new StringBuilder();
	    		for(int j=first;j<=last;j++)
				{
					row.append(words[j]);
					if(j<last)
					{
						for(int i=0;i<countkongge;i++)
							row.append(" ");
						if(isjunfen>0) //不可以均分 ，多出的空格依次放在左边
							row.append(" ");
						isjunfen--;
					}
					
				}
				for(int j=row.length();j<L;j++)  
	            {  
	                row.append(" ");  
	            }    
    			res.add(row.toString());
    			first=last+1;
    			count=0;	

	    	}else{
	    		count+=words[last].length();
	    	}
	    }
	    //单独处理最后一行.因为最后一行的空格分布与前面不一样
	    StringBuilder row=new StringBuilder();
	    for(int i=first;i<words.length;i++)
	    {
	    	row.append(words[i]); 
	    	if(row.length()<L)
	    		row.append(" ");
	    }
	    for(int i=row.length();i<L;i++)  
	    {  
	        row.append(" ");  
	    }  
	    res.add(row.toString());
	    return res;    	
    }
}

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