这题有点意思,首先想到分两步处理,一是平均化单词(就是使空格符平均分布,其实也简单,就是从第一个空格符起,一个一个的插入,下一次的插入位置是上一次最后的空格后一个位置,到结尾时再次从头开始,直到不能再放了),一是分解单词(就是往里放单词,这部要算上空格的数量,直到不能放为止。),因为最后一行的处理方式不一样,所以单独一个函数处理最后一行;针对一个单词的情况还要优化否则的话,时间超时。
// LeetCode_TextJustification.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include <iostream>
#include <vector>
#include <string>
using namespace std;
string lastAnwserLeft(vector<string> &words, int L) {
string ret="";
int len = words.size();
int j=0;
for(int i=0;i<len;i++)
{
ret += words[i];
j += words[i].length();
if (j<L&&words[i]!="")
{
ret += " ";
j += 1;
}
}
while(j<L)
{
ret += " ";
j++;
}
return ret;
}
string OneAnwserEven(vector<string> &words, int L) {
string ret="";
int len = words.size();
if (len==1)// this is necessary otherwise overtime
{
ret += words[0];
int j = words[0].length();
while(j<L)
{
ret += " ";
j++;
}
return ret;
}
int i,j=0;
for (i=0;i<len-1;i++)
{
ret += words[i];
j += words[i].length();
ret += " ";
j += 1;
}
ret += words[i];
j += words[i].length();
string::size_type lastempty=0,curempty;
while(j<L)
{
curempty = ret.find(' ',lastempty);//find return string::npos when failure
if(curempty>=ret.length())
{
lastempty = 0;
continue;
}
ret.insert(curempty,1,' ');
j++;
while(ret[curempty]==' ')
curempty++;
lastempty = curempty;
}
return ret;
}
vector<string> fullJustify(vector<string> &words, int L) {
vector<string> ret;
vector<string> onegroup;
string oneanswer;
int lenwords = words.size();
int i = 0,j=0;
onegroup.clear();
while(i<lenwords)
{
while(j+onegroup.size()<=L+1)//j+onegroup.size()-1<=L cannot work
{
onegroup.push_back(words[i]);
j += words[i].length();
i++;
if (i>=lenwords)
{
break;
}
}
if (j+onegroup.size()>L+1)
{
j -= onegroup[onegroup.size()-1].length();
onegroup.pop_back();
i--;
oneanswer = OneAnwserEven(onegroup,L);
ret.push_back(oneanswer);
oneanswer.clear();
onegroup.clear();
j=0;
}
else
{
oneanswer = lastAnwserLeft(onegroup,L);
ret.push_back(oneanswer);
oneanswer.clear();
onegroup.clear();
j=0;//break;//
}
}
return ret;
}
int _tmain(int argc, _TCHAR* argv[])
{
//string ret;
vector<string> ret;
vector<string> words;
words.push_back("This");
words.push_back("is");
//words.push_back("the");
words.push_back("an");
words.push_back("example");
words.push_back("of");
words.push_back("text");
words.push_back("justification.");
//ret = OneAnwserEven(words,16);
ret = fullJustify(words,16);
for (int i=0;i<ret.size();i++)
{
cout<<ret[i]<<endl;
}
//cout<<ret<<endl;
system("pause");
return 0;
}