力扣每日一题20210909文本左右对齐

题目描述

文本左右对齐

思路&＃xff1a;模拟

统计构成一行的单词后&＃xff0c;将该行进行左右对齐的空格补齐。

class Solution:def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:# 将word[left]到word[right-1]的单词组合成一个左右对齐的行def process(left, right):# 结尾行&＃xff0c;向后补齐空格if right &＃61;&＃61; n:res &＃61; &＃39; &＃39;.join(words[left:right])res &＃43;&＃61; &＃39; &＃39; * (maxWidth - len(res))return resspaces &＃61; right - left - 1# 单独单词构成的行&＃xff0c;向后补齐空格if not spaces:return words[left] &＃43; &＃39; &＃39; * (maxWidth - len(words[left]))cur &＃61; sum(len(w) for w in words[left:right]) &＃43; spaces# 统计每个空隙间的空格数each &＃61; [1] * spacesj &＃61; 0while cur < maxWidth:# 根据题意优先补左边&＃xff0c;所以从0开始each[j] &＃43;&＃61; 1j &＃43;&＃61; 1if j &＃61;&＃61; spaces:j &＃61; 0cur &＃43;&＃61; 1j &＃61; 0res &＃61; &＃39;&＃39;# 根据每个间隙的空格数构建该行字符串while left < right:res &＃43;&＃61; words[left]if left < right - 1:res &＃43;&＃61; &＃39; &＃39; * each[j]j &＃43;&＃61; 1left &＃43;&＃61; 1return resn &＃61; len(words)idx &＃61; i &＃61; 0ans &＃61; []while idx < n:i &＃61; idxcurLen &＃61; len(words[idx])idx &＃43;&＃61; 1# 统计哪些单词组成一行while idx < n and curLen < maxWidth:# 单词之间需要空格curLen &＃43;&＃61; 1curLen &＃43;&＃61; len(words[idx])idx &＃43;&＃61; 1if curLen > maxWidth:idx -&＃61; 1curLen -&＃61; len(words[idx]) &＃43; 1ans.append(process(i, idx))return ans