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Leetcode 68 Text Justification

來源：程序員人生發布時間：2016-11-16 08:11:12 閱讀次數：2585次

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.

大摹擬，寫了很冗雜的代碼，wa了無數次終究過了，后來看到了簡介的版本，對側重寫了1份，詳見注釋！

class Solution { public: vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> result; for(int i=0,k,l;i<words.size();i+=k) { for(k=l=0;i+k<words.size() && l+words[i+k].size()<=maxWidth-k;k++) l+=words[i+k].size(); //找出1行單詞的個數k和總字母數l string temp=words[i]; //將第1個單詞放入 for(int j=0;j<k⑴;j++) { if(i+k>=words.size()) //如果是最后1行 temp+=' '; else temp+=string((maxWidth-l)/(k⑴)+(j<(maxWidth-l)%(k⑴)),' ');//精華：單詞之間的空格 temp+=words[i+j+1]; } temp+=string(maxWidth-temp.size(),' ');//補全空格 result.push_back(temp); } return result; } };

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Leetcode 68 Text Justification