Text Justification_justification学术-优快云博客

本文链接：https://blog.youkuaiyun.com/pkufergus/article/details/11826093

本文介绍了一种文本格式化算法，该算法能够将给定的单词数组按指定长度进行格式化处理，确保每行文字都刚好达到指定长度，并实现左右对齐效果。文章详细解释了如何贪婪地填充单词，均匀分布空格，并处理最后一行的左对齐情况。

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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.

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.

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int space = 0;
		int i, j;
		vector<string> ans;
		string line;
		int size = words.size();
		int len = 0;
		int k;
		if(size == 1){
			line = words[0];
			line += string(L - words[0].size(), ' ');
			ans.push_back(line);
			return ans;
		}
		for(i = 0; i < size; i++){
			len = 0;
			line = "";
			for(j = i; j < size; j++){
				if(len + words[j].size() > L){
					space = L + 1 - len;
					break;
				} else if(len + words[j].size() == L){
					space = 0;
					break;
				}
				len += words[j].size();
				len += 1;
			}
			if(j == size){
				for(k = i; k < j - 1; k++){
					line+=words[k];
					line+=" ";
				}
				line+=words[k];
				line += string(L - line.size(), ' ');
				ans.push_back(line);
				break;
			} else if(space == 0){
				for(k = i; k < j; k++){
					line+=words[k];
					line+=" ";
				}
				line+=words[k];
				ans.push_back(line);
				i = j;
			} else {
				if(j - 1 == i){
					line += words[i];
					line += string(L - words[i].size(), ' ');
					ans.push_back(line);
				} else {
					int c = space /(j - 1 - i);
					int d = space % (j - 1 - i);
					for(k = i; k < j - 1; k++){
						line += words[k];
						if(k - i < d){
							line += string(c + 2, ' ');
						} else {
							line += string(c + 1, ' ');
						}
					}
					line+=words[k];
					ans.push_back(line);
				}
				i = j - 1;
			}
		}
		return ans;
    }
};