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 "
]#
# @lc app=leetcode.cn id=68 lang=python3
#
# [68] 文本左右对齐
## @lc code=start
class Solution:def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:n = len(words)start, end = 0, n-1row = 0ans = [[' ' for _ in range(maxWidth)] for _ in range(n)]while(start<n):total = 0num = 0pos = startwhile(pos<n):flag=Falseif total + len(words[pos]) + num>maxWidth:flag=Truebreaktotal+=len(words[pos])num+=1pos+=1if flag:total=maxWidth-totalspace,r=divmod(total,max(1,num-1))pp=0for i in range(start,start+num):ans[row][pp:pp+len(words[i])] = words[i]pp=pp+len(words[i])+spaceif r>0:pp+=1r-=1else:pp=0for i in range(start,start+num):ans[row][pp:pp+len(words[i])] = words[i]pp=pp+len(words[i])+1breakstart = posrow+=1tmp = []for i in range(row+1):tmp.append(''.join(ans[i]))return tmp# @lc code=end
