LeetCode——Text Justification

最新推荐文章于 2025-12-06 07:43:37 发布

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本文介绍了一个文本格式化问题，即如何将给定的单词数组按照指定长度L进行排版，确保每行恰好包含L个字符，并实现左对齐或两端对齐的效果。详细解释了处理逻辑，包括对最后一行的特殊处理。

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.

» Solve this problem

题目不难，就是字符串的处理。

把逻辑理清楚：

1.对每一行，放尽量多的单词，长度不能超过L，用尽可能平均的空格分隔单词；

2.对最后一行，用一个空格分隔单词，最后剩余部分用空格填充。

class Solution {
private:
    string handle_normal(vector<string> &word, int L, int wordL) {
        string answer;
        if (word.size() == 1) {
    		answer += word[0];
    		for (int i = 0; i < L - wordL; i++) {
    			answer += " ";
    		}
    		return answer;
    	}
        int d = (L - wordL) / (word.size() - 1);
        int r = (L - wordL) % (word.size() - 1);
        answer = word[0];
        for (int i = 1; i < word.size(); i++) {
            for (int j = 0; j < d; j++) {
                answer += " ";
            }
            if (r > 0) {
                answer += " ";
                r--;
            }
            answer += word[i];
        }
        return answer;
    }
    
    string handle_end(vector<string> &word, int L, int count) {
        string answer;
        answer = word[0];
        for (int i = 1; i < word.size(); i++) {
            answer += " " + word[i];
        }
        for (int i = 0; i < L - count; i++) {
            answer += " ";
        }
        return answer;
    }
    
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> ans;
        vector<string> selWord;
        int count = -1, wordL = 0;
        for (vector<string>::iterator iter = words.begin(); iter != words.end(); iter++) {
            if (count + iter->length() + 1 > L) {
                ans.push_back(handle_normal(selWord, L, wordL));
                selWord.clear();
                count = -1;
                wordL = 0;
            }
            count += 1 + iter->length();
            wordL += iter->length();
            selWord.push_back(*iter);
        }
        ans.push_back(handle_end(selWord, L, count));
        return ans;
    }
};