How do I handle words of different lengths in Print Words Vertically?

Use the length of the longest word to define columns and add spaces for shorter words, trimming trailing spaces afterward.

Can this problem be solved without using nested loops?

Nested iteration over word indices and word array is the simplest approach; alternative methods still mimic this column-by-column access.

Why are trailing spaces a concern in vertical word printing?

Trailing spaces are not allowed by the problem constraints; forgetting to trim them results in failed test cases.

Does the solution rely on any specific data structures?

A simple array of strings suffices, leveraging index access for array plus string manipulation.

Is this problem pattern commonly seen in interviews?

Yes, it tests array and string manipulation with simulation patterns, specifically handling vertical alignment of character sequences.

#1324

Medium

auto_awesome数组·string

LeetCode 题解工作台

竖直打印单词

给你一个字符串 s 。请你按照单词在 s 中的出现顺序将它们全部竖直返回。单词应该以字符串列表的形式返回，必要时用空格补位，但输出尾部的空格需要删除（不允许尾随空格）。每个单词只能放在一列上，每一列中也只能有一个单词。示例 1：输入： s = "HOW ARE YOU" 输出： ["HAY"…

数组字符串模拟

题目描述

给你一个字符串 s。请你按照单词在 s 中的出现顺序将它们全部竖直返回。
单词应该以字符串列表的形式返回，必要时用空格补位，但输出尾部的空格需要删除（不允许尾随空格）。
每个单词只能放在一列上，每一列中也只能有一个单词。

示例 1：

输入：s = "HOW ARE YOU"
输出：["HAY","ORO","WEU"]
解释：每个单词都应该竖直打印。 
 "HAY"
 "ORO"
 "WEU"

示例 2：

输入：s = "TO BE OR NOT TO BE"
输出：["TBONTB","OEROOE","   T"]
解释：题目允许使用空格补位，但不允许输出末尾出现空格。
"TBONTB"
"OEROOE"
"   T"

示例 3：

输入：s = "CONTEST IS COMING"
输出：["CIC","OSO","N M","T I","E N","S G","T"]

提示：

1 <= s.length <= 200
s 仅含大写英文字母。
题目数据保证两个单词之间只有一个空格。

lightbulb

解题思路

方法一：模拟

我们先将字符串 $s$ 按空格分割成单词数组 $words$ ，然后遍历单词数组，找出最长的单词长度 $n$ 。

接下来我们从第 $1$ 到第 $n$ 个字符，分别从单词数组中取出对应的字符，如果当前单词长度不足，则用空格补齐，放到一个字符串 $t$ 中。最后将 $t$ 去掉末尾的空格，加入答案数组中即可。

时间复杂度 $O(m)$ ，空间复杂度 $O(m)$ 。其中 $m$ 为字符串 $s$ 的长度。

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class Solution:
    def printVertically(self, s: str) -> List[str]:
        words = s.split()
        n = max(len(w) for w in words)
        ans = []
        for j in range(n):
            t = [w[j] if j < len(w) else ' ' for w in words]
            while t[-1] == ' ':
                t.pop()
            ans.append(''.join(t))
        return ans

speed

复杂度分析

指标	值
时间	complexity is O(n * m), where n is the number of words and m is the length of the longest word, because each character in each word is visited once. Space complexity is also O(n * m) due to storing vertical strings in the result array.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Are you considering varying word lengths for column generation?
question_mark
How do you handle trailing spaces when shorter words end before longer ones?
question_mark
Can you simulate the vertical printing without using extra nested arrays?

warning

常见陷阱

外企场景

error
Not trimming trailing spaces, resulting in incorrect output format.
error
Assuming all words have equal length, causing index out-of-range errors.
error
Appending characters in the wrong order and losing the original word sequence.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return words rotated diagonally instead of strictly vertical columns.
arrow_right_alt
Handle lowercase and mixed-case letters while preserving original formatting.
arrow_right_alt
Support multiple spaces between words and preserve spacing alignment in output.

help

常见问题

外企场景

继续练习

#2011 执行操作后的变量值 #2056 棋盘上有效移动组合的数目 #2109 向字符串添加空格