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."]
    L16.

    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.
    贪心,注意细节。
     1 class Solution {
     2 private:
     3     vector<string> ret;
     4 public:
     5     vector<string> fullJustify(vector<string> &words, int L) {
     6         // Start typing your C/C++ solution below
     7         // DO NOT write int main() function
     8         ret.clear();
     9         int i = 0;
    10         while(i < words.size())
    11         {
    12             int size = 0;
    13             int beg = i;
    14             while(i < words.size())
    15             {
    16                 int newSize = size == 0 ? words[i].size() : size + 1 + words[i].size();
    17                 if (newSize <= L)
    18                     size = newSize;
    19                 else
    20                     break;
    21                 i++;
    22             }
    23             
    24             int spaceCount = L - size;
    25             int everyCount;
    26             if (i - 1 - beg != 0 && i != words.size())
    27             {
    28                 everyCount = spaceCount / (i - 1 - beg);
    29                 spaceCount %= (i - 1 - beg);
    30             }
    31             else
    32                 everyCount = 0;
    33             string s;
    34             for(int j = beg; j < i; j++)
    35                 if (j == beg)
    36                     s = words[j];
    37                 else
    38                 {
    39                     s += ' ';
    40                     for(int k = 0; k < everyCount; k++)
    41                         s += ' ';
    42                     if (spaceCount > 0 && i != words.size())
    43                     {
    44                         s += ' ';
    45                         spaceCount--;
    46                     }
    47                     s += words[j];
    48                 }
    49                 
    50             for(int j = 0; j < spaceCount; j++)
    51                 s += ' ';
    52                 
    53             ret.push_back(s);
    54         }
    55         
    56         return ret;
    57     }
    58 };
  • 相关阅读:
    十步完全理解SQL
    c#退出应用程序办法
    几个有意思的算法题
    GeoServer不同服务器安装配置、数据发布及客户端访问
    开启httpd服务的时候 显示Could not reliably determine the server`s fully qualified domain name
    Working With OpenLayers(Section 1: Creating a Basic Map)
    GeoServer地图开发解决方案(五):基于Silverlight技术的地图客户端实现
    模拟远程HTTP的POST请求
    模拟提交带附件的表单
    支付宝手机网站接口对接
  • 原文地址:https://www.cnblogs.com/chkkch/p/2776335.html
Copyright © 2011-2022 走看看