题目：
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.
代码：
class Solution {
public:
vector<string> fullJustify(vector<string>& words, int maxWidth) {
vector<string> ret;
vector<string> tmp; // words in one line
int width = 0; // total width in one line
const char BLANK=‘ ‘;
int i=0;
while ( i<words.size() )
{
// judge if a word can be added in a line
if ( width+words[i].size() <= maxWidth )
{
if ( width+words[i].size()==maxWidth )
{
tmp.push_back(words[i]);
width = maxWidth;
}
else
{
tmp.push_back(words[i]+string(1,BLANK));
width += words[i].size()+1;
}
}
// or create a new line & handle the words[i]
else
{
// CREAT A NEW LINE
// number of blank need to add
int blank_num = maxWidth - width;
if ( tmp.back()[tmp.back().size()-1]==BLANK )
{
tmp.back() = string(tmp.back().begin(),tmp.back().end()-1);
blank_num = blank_num + 1;
}
// number of position for blanks
int pos_num = tmp.size()-1;
if ( pos_num==0 )
{
ret.push_back(tmp.back()+string(blank_num, BLANK));
}
else
{
// number of blanks remain to be evenly distributed
int blank_remain = blank_num % pos_num;
for ( int j=0; j<tmp.size()-1; ++j )
{
tmp[j] = tmp[j]+ string(blank_num/pos_num, BLANK);
}
for ( int j=0; j<blank_remain; ++j )
{
tmp[j] = tmp[j] + string(1, BLANK);
}
string str = "";
for ( int j=0; j<tmp.size(); ++j ) str += tmp[j];
ret.push_back(str);
}
// HANDLE THE words[i]
tmp.clear();
if ( words[i].size()==maxWidth )
{
tmp.push_back(words[i]);
width = maxWidth;
}
else
{
tmp.push_back(words[i]+string(1, BLANK));
width = words[i].size()+1;
}
}
++i;
}
// address the remain line
if ( tmp.size()!=0 )
{
int blank_num = maxWidth - width;
string last_line = "";
for ( int i=0; i<tmp.size(); ++i ) last_line += tmp[i];
ret.push_back(last_line+string(blank_num, BLANK));
}
return ret;
}
};
tips：
1. 要对string的各种操作都很熟悉
2. 要理解对题意，重点是到底啥是evenly：意思是不能evenly，就从左边到右一个个分配。
3. 扫一扫各种test case，多扫几遍可以AC了。
这道题自己扫了4次，终于AC了；经验就是，如果不能做到第一次把所有关键细节都考虑完全了，至少把一些细节（诸如blank_num，pos_num，blank_remain）单独列拎出来，这样做的好处就是如果遇上各种需要考虑的corner cases可以单独处理这些关键细节，而不用影响其他的部分。