[LeetCode] 68. Text Justification 文本对齐
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 Lcharacters.
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.
将由单词组成的数组,按每行最多L个字符进行对齐调整。
解法:先找到某一行能放下的单词,单词的字母数之和加上单词之间的空格要小于等于L,然后需要补充空格的话在填进去空格。假设有n个单词,那么就会有n-1个间隔,每个间隔的空格数应该是:(L-此行所有单词的长度和)/(n-1)。只用一个单词的话,排好单词在补上空格即可。
参考:喜刷刷
Java:
public List<String> fullJustify(String[] words, int L) { List<String> res = new ArrayList<>(); int n = words.length; char[] spaces = new char[L]; Arrays.fill(spaces, ' '); for(int i=0; i<n; i++) { int len = words[i].length(); int j = i; while(i<n-1 && len+1+words[i+1].length()<=L) { len += 1+words[++i].length(); } StringBuilder sb = new StringBuilder(words[j]); if(j == i || i==n-1) { while(i==n-1 && j < i) { sb.append(" "+words[++j]); } sb.append(spaces, 0, L-sb.length()); } else { int avg = (L-len)/(i-j); int rem = (L-len)%(i-j); while(j < i) { sb.append(spaces, 0, avg+1); if(rem-- > 0) sb.append(" "); sb.append(words[++j]); } } res.add(sb.toString()); } return res; }
Python:
class Solution(object): def fullJustify(self, words, maxWidth): """ :type words: List[str] :type maxWidth: int :rtype: List[str] """ def addSpaces(i, spaceCnt, maxWidth, is_last): if i < spaceCnt: # For the last line of text, it should be left justified, # and no extra space is inserted between words. return 1 if is_last else (maxWidth // spaceCnt) + int(i < maxWidth % spaceCnt) return 0 def connect(words, maxWidth, begin, end, length, is_last): s = [] # The extra space O(k) is spent here. n = end - begin for i in xrange(n): s += words[begin + i], s += ' ' * addSpaces(i, n - 1, maxWidth - length, is_last), # For only one word in a line. line = "".join(s) if len(line) < maxWidth: line += ' ' * (maxWidth - len(line)) return line res = [] begin, length = 0, 0 for i in xrange(len(words)): if length + len(words[i]) + (i - begin) > maxWidth: res += connect(words, maxWidth, begin, i, length, False), begin, length = i, 0 length += len(words[i]) # Last line. res += connect(words, maxWidth, begin, len(words), length, True), return res
C++:
class Solution { public: vector<string> fullJustify(vector<string>& words, int maxWidth) { int cnt = 0, left = 0; vector<string> result; for(int i =0; i< words.size(); i++) { cnt += words[i].size(); if(cnt+i-left > maxWidth || i+1==words.size()) { if(cnt+i-left > maxWidth) cnt -= words[i--].size(); string str = words[left]; for(int j = left+1; j<= i; j++) { int m = maxWidth-cnt, n = i-left; if(i+1==words.size()) str += " "; else str.append(m/n + (j-left-1<m % n), ' '); str += words[j]; } str.append(maxWidth-str.size(), ' '); result.push_back(str); left = i+1, cnt = 0; } } return result; } };
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