LeetCode Text Justification

最新推荐文章于 2020-06-10 05:27:24 发布

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本文介绍了一个用于文本格式化的算法，该算法能够确保每行文本恰好包含指定数量的字符，并实现左右两端对齐的效果。通过贪婪策略填充尽可能多的单词到一行中，并在必要时添加额外的空格来达到规定的长度。

题目：

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.

class Solution {
public:
	vector<string> fullJustify(vector<string> &words, int L) {
		vector<string> ans;
		int beg = 0;
		while (beg < words.size()) {
			int i = beg;
			int wordCnt = 0, total = 0;
			string s;
			for (; i < words.size(); i++) {
				if (total <= L) {
					if (wordCnt == 0)
						total += words[i].size();
					else
						total += words[i].size() + 1;
					wordCnt++;
					if (total > L)
						break;
				}
			}
			if (total > L) {
				total -= (words[i].size()+1);
				wordCnt--;
			}
			//对每行进行处理，每行只有一个单词
			if (wordCnt == 1) {
				int space = L - total;
				s.insert(s.begin(), words[beg].begin(), words[beg].end());
				s.insert(s.end(), space, ' ');
			}
			//最后一行特殊处理
			else if (i == words.size()) {
				int space = L - total;
				s.insert(s.begin(), words[beg].begin(), words[beg].end());
				for (int j = beg + 1; j < beg + wordCnt; j++) {
					s.insert(s.end(), 1, ' ');
					s.insert(s.end(), words[j].begin(), words[j].end());
				}
				s.insert(s.end(), space, ' ');
			}
			else {
				int space2 = (L - total) / (wordCnt - 1);
				//前pre个的空格为space1个，之后的空格为space2
				int pre = (L - total) % (wordCnt - 1);
				int space1 = space2 + 1;
				s.insert(s.begin(), words[beg].begin(), words[beg].end());
				for (int j = beg + 1; j < beg + wordCnt; j++) {
					//插入空格
					if (pre > 0) {
						s.insert(s.end(), space1+1, ' ');
						pre--;
					}
					else
						s.insert(s.end(), space2+1, ' ');
					s.insert(s.end(), words[j].begin(), words[j].end());
				}
			}
			ans.push_back(s);
			beg = i;
		}
		return ans;
	}
};