zoukankan      html  css  js  c++  java
  • LeetCode_Text Justification

    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.
    

      分析: 主要注意两个地方。一, 正常的句子单词之间的空格数目是一;二,最后一行句子按正常处理

    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> res;
            if(words.size() < 1) return res;
            
            int start , end, len , evenSpace, bonus;
            
            end = 0;
            while(true){
            
                len = 0;
                for(start = end; end < words.size(); ++end){
                    if(len +(end - start) + words[end].length() > L) break;// normal sentence has one space between words
                    len += words[end].length();    
                }
                end--;
                
                if(end == start){ // only one word
                    string line = words[end];
                    line.append(L - len, ' ');
                    res.push_back(line);
                }else{ // multiply words
                    
                    evenSpace = (L - len)/(end - start);
                    bonus = L -len - evenSpace*(end - start);
                    bool isLastLine = (end == words.size() -1);
                    if(isLastLine){
                        evenSpace = 1;
                        bonus = 0;
                    }
                    
                    string line = words[start];
                    for(int i = start + 1; i <= end; ++i)
                    {
                        int space = evenSpace;
                        if(bonus > 0)
                        {
                            --bonus;
                            ++space;
                        }
                        line.append(space,' ');
                        line.append(words[i]);
                    }
                    
                    if(isLastLine){
                        line.append(L-len - (end - start), ' ');
                    }
                    
                    res.push_back(line);
                }
                
                end++;
                if(end == words.size()) break;
            }
            
            return res;        
        }
    };
  • 相关阅读:
    关于时间的字词
    Postgresql 存储过程调试 1
    Delphi 调试日子
    Delphi 调试日子
    TList,TObjectList 使用——资源释放
    Lazarus开发环境编译选项配置
    Delphi 递归搜索.SVN文件夹并“处理”
    Delphi 路径相关函数
    如何掌握程序语言(王垠)
    struct/class等内存字节对齐问题详解
  • 原文地址:https://www.cnblogs.com/graph/p/3290497.html
Copyright © 2011-2022 走看看