Text Justification

最新推荐文章于 2024-03-15 13:24:07 发布

原创最新推荐文章于 2024-03-15 13:24:07 发布 · 474 阅读

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本文介绍了一种方法，用于将单词数组排列成指定长度的行，并进行左右对齐。通过贪婪策略填充额外的空格，确保每行恰好包含指定数量的字符。详细解释了如何在不同行间均匀分配空格，特别是对于最后一行的独特处理。

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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.

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.

解法：考虑下边界情况就行了

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> ret;
        int last = 0;
        if(words.size() > 0) addStr(words,L,ret,0,last);
        
        return ret;
    }
    
    void addStr(vector<string> &words,int L,vector<string>& ret,int cur,int& last)
    {
        if(cur >= words.size() ||  words[cur].length() > L) return;
        
        string str = "";
        int total_len = 0;  
        vector<int> index;
        
        total_len += words[cur].length();
        index.push_back(cur);	
        if(cur == words.size()-1){
            last = 1;    
        }else{
            cur++;
            while(cur < words.size() && total_len + words[cur].length()+1 <= L )
            {
                if(cur == words.size()-1){
                    last = 1;    
                }
            	total_len += (words[cur].length()+1);
                index.push_back(cur++);
            }
        }
    
        total_len -= (index.size()-1);
 
        if(last == 1 || index.size() == 1){
            for(int i = 0;i< index.size()-1;i++)
            {
                str.append(words[index[i]]+" ");
            }  
            str.append(words[index[index.size()-1]]);
            str.append(L - total_len - index.size()+1,' ');
        }else{
            int aver = (L - total_len)/(index.size()-1);
            int extra = (L - total_len) - aver*(index.size()-1);
            
            for(int i = 0;i<index.size()-1;i++)
            {
                str.append(words[index[i]]);
                str.append(aver,' ');
                if(i < extra) str.append(" ");
            }
            str.append(words[index[index.size()-1]]);
        }
        ret.push_back(str);
        if(last == 1) return;
        else addStr(words,L,ret,cur,last);
    }
};

12 milli secs.