68. 文本左右对齐
给定一个单词数组和一个长度 maxWidth,重新排版单词,使其成为每行恰好有 maxWidth 个字符,且左右两端对齐的文本。
你应该使用“贪心算法”来放置给定的单词;也就是说,尽可能多地往每行中放置单词。必要时可用空格 ' ' 填充,使得每行恰好有 maxWidth 个字符。
要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配,则左侧放置的空格数要多于右侧的空格数。
文本的最后一行应为左对齐,且单词之间不插入额外的空格。
说明:
单词是指由非空格字符组成的字符序列。
每个单词的长度大于 0,小于等于 maxWidth。
输入单词数组 words 至少包含一个单词。
示例:
输入:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
输出:
[
"This is an",
"example of text",
"justification. "
]
示例 2:
输入:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
输出:
[
"What must be",
"acknowledgment ",
"shall be "
]
解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
因为最后一行应为左对齐,而不是左右两端对齐。
第二行同样为左对齐,这是因为这行只包含一个单词。
示例 3:
输入:
words = ["Science","is","what","we","understand","well","enough","to","explain",
"to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
输出:
[
"Science is what we",
"understand well",
"enough to explain to",
"a computer. Art is",
"everything else we",
"do "
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/text-justification
class Solution: def fullJustify(self, words: List[str], maxWidth: int) -> List[str]: res=[] i=0 while i < len(words): ii=i l=len(words[i]) j=1#这一行的单词个数 while i<len(words)-1 and l<=maxWidth:#贪心找最多,以相邻单词间隔1空格为标准 i+=1 j+=1 l=l+1+len(words[i]) d=maxWidth-l if d<0:#超过maxWidth减去最后一个单词 i-=1 j-=1 l=l-len(words[i+1])-1 d=maxWidth-l if j==1: w=words[ii]+' '*(maxWidth-len(words[ii])) elif i==len(words)-1:#最后一行为左对齐 w='' for k in range(ii,ii+j): if k<ii+j-1: w+=words[k]+' ' else: w+=words[k] w+=' '*(maxWidth-len(w)) else: if d>0:#除相邻单词间隔1空格外还有剩余空格 if d%(j-1)==0:#剩余空格可均分 dd=d//(j-1) w='' for k in range(ii,ii+j): if k<ii+j-1: w+=words[k]+' '*(dd+1) else: w+=words[k] else:#剩余空格不可均分 dd=d//(j-1) ddd=d%(j-1) w='' for k in range(ii,ii+j): if k<ii+ddd: w+=words[k]+' '*(dd+2)#均分后多余的往前面加 elif k<ii+j-1: w+=words[k]+' '*(dd+1) else: w+=words[k] else:#相邻单词刚好间隔1空格 w='' for k in range(ii,ii+j): if k<ii+j-1: w+=words[k]+' ' else: w+=words[k] res.append(w) i+=1 return res