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[leetcode]Text Justification
问题描述;
Given an array of words and a length L, format the text such that each line has exactlyL 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 exceedL in length.
此题写的不好,有些复杂了,代码写的也是一陀。。
代码:
vector<string> fullJustify(vector<string> &words, int L) { //C++ vector<string> result; int size = words.size(); int num = 0; string newline = ""; vector<string> line; for(int i=0; i<size; i++){ int left = L-num; int len = words[i].size(); if(left >= len){ num +=len+1; line.push_back(words[i]); } else { int space = L-num+line.size(); for(int j=0; j<line.size()-1; j++){ int tempNum = ceil((double)space/(line.size()-1-j)); newline +=line[j]+string(tempNum,' '); space -=tempNum; } newline +=line[line.size()-1]; if(space >0) newline += string(space,' '); result.push_back(newline); newline =""; line.clear(); num=0; i--; } } for(int i=0; i<line.size()-1; i++) newline+=line[i]+" "; newline += line[line.size()-1]; if(L-newline.size() >0) newline += string(L-newline.size(),' '); result.push_back(newline); return result; }
[leetcode]Text Justification