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Leetcode 68 Text Justification

来源：程序员人生发布时间：2016-11-16 08:11:12 阅读次数：2136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.

大摹拟，写了很冗杂的代码，wa了无数次终究过了，后来看到了简介的版本，对侧重写了1份，详见注释！

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行单词的个数k和总字母数l string temp=words[i]; //将第1个单词放入 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)),' ');//精华：单词之间的空格 temp+=words[i+j+1]; } temp+=string(maxWidth-temp.size(),' ');//补全空格 result.push_back(temp); } return result; } };

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