【LeetCode】68. Text Justification

问题描述

https://leetcode.com/problems/text-justification/#/description

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.

算法

文本显示调整问题，将一系列单词words[n]按行显示，每一行做多只能有L个字母，如：
words=["This", "is", "an", "example", "of", "text", "justification."]，L=16

最终显示的行为：["This is an", "example of text", "justification. "]

输出的行需要满足几个条件

每一行的单词要尽可能的多，并且要左右都能顶格，不足L个字符时，可以使用空格' '代替字母
每一行单词之间的空格要尽可能地均匀，不能满足均匀的前提下，左边的要比右边的多
最后一行只要满足左边顶格，并且单词之间不要有多余空格

算法：

遍历到一个单词时，往前探到几个单词符合要求，即单词的字母数之和加上单词之间的空格要小于等于L，探到符合要求的单词数后就开始组织行。

因为单词已经有了，接下来就是安排好单词的位置了，有n个单词，那么就会有n-1个间隔，那么除了最后一行外的每个间隔的空格数应该是：(L-n个单词的字母总量)/(n-1)，但因为有时不能除尽，最后一个间隔有可能会少，要进行判断。最后一行按照普通的做法，从左到右排好单词即可。

参考自：https://discuss.leetcode.com/topic/9147/simple-java-solution

代码

        public List<String> fullJustify(String[] words, int maxWidth) {
	        List<String> lines = new ArrayList<>();
	        int cur = 0;
	        while(cur < words.length) {
	        	int last = cur+1;
	        	int cnt = words[cur].length(); // 单词的字母数
	        	while(last < words.length) {
	        		if(words[last].length() + cnt + 1 > maxWidth) {
	        			break;
	        		}
	        		cnt += words[last].length()+1;
	        		last++;
	        	}
	        	int gap = last - cur - 1;
	        	StringBuilder sb = new StringBuilder();
	        	if(last == words.length || gap == 0) { // 到了最后一行或者只有一个单词
	        		for(int i=cur;i<last;i++) {
	        			sb.append(words[i]+" ");
	        		}
	        		sb.deleteCharAt(sb.length()-1);
	        		for(int i=sb.length();i<maxWidth;i++) {
	        			sb.append(" ");
	        		}
	        	} else {
	        		int spaces = (maxWidth - cnt) / gap;
	        		int rem = (maxWidth - cnt) % gap;
	        		for(int i=cur;i<last;i++) {
	        			sb.append(words[i]);
	        			if(i<last-1) {
	        				for(int j=0;j<=(spaces+(i-cur<rem?1:0));j++) {
	        					sb.append(" ");
	        				}
	        			}
	        		}
	        	}
	        	lines.add(sb.toString());
	        	cur = last;
	        }
	        return lines;
	    }