LeetCode_OJ【68】Text Justification

Given an array of words and a length L, format the text such that each line has exactlyL 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.

这题实现起来还是比较繁琐的，要注意的地方比较多。

题目最后说每个word都能保证长度不超过给定的L ,这样我们就不用考虑非法输入了。

另外最后一行和只有一个word的行要特殊处理。

java实现如下：

public class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> res = new ArrayList<String>();
for(int i = 0,len ,start,wordSum ; i < words.length ; ){
len = 0;
wordSum = 0;
start = i;
StringBuilder line = new StringBuilder();
while(i < words.length && len + words[i].length() <= maxWidth){
wordSum += words[i].length();
len += (1 + words[i].length()); // 默认包含最少一个空格
i++ ;
}
int count = i - start ; //该行word个数
if(count != 1 && i != words.length ){ //i取到length时说明是最后一行
int space = (maxWidth - wordSum) / (count-1);
int rest = (maxWidth - wordSum) % (count-1);
for(int m = 0 ; m < count ; m ++){
line.append(words[start + m]);
//最后一个单词不加空格
if(m != count-1){
for(int n = 0 ; n < space ; n ++)
line.append(" ");
if(rest -- > 0)
line.append(" ");
}
}
}
else {
while(count -- > 0){
line.append(words[start ++]);
if(count != 0)
line.append(" ");
}
while(line.length() < maxWidth)
line.append(" ");
}
res.add(line.toString());
}
return res;
}
}