solution.md 2.4 KB
Newer Older
每日一练社区's avatar
每日一练社区 已提交
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123
# Z 字形变换

<div class="notranslate">
    <p>将一个给定字符串 <code>s</code> 根据给定的行数 <code>numRows</code> ,以从上往下、从左到右进行&nbsp;Z 字形排列。</p>

    <p>比如输入字符串为 <code>"PAYPALISHIRING"</code>&nbsp;行数为 <code>3</code> 时,排列如下:</p>

    <pre>
    P   A   H   N
    A P L S I I G
    Y   I   R</pre>

    <p>之后,你的输出需要从左往右逐行读取,产生出一个新的字符串,比如:<code>"PAHNAPLSIIGYIR"</code>。</p>

    <p>请你实现这个将字符串进行指定行数变换的函数:</p>

    <pre>string convert(string s, int numRows);</pre>

    <p>&nbsp;</p>

    <p><strong>示例 1:</strong></p>

    <pre><strong>输入:</strong>s = "PAYPALISHIRING", numRows = 3
<strong>输出:</strong>"PAHNAPLSIIGYIR"
</pre>
    <strong>示例 2:</strong>

    <pre><strong>输入:</strong>s = "PAYPALISHIRING", numRows = 4
<strong>输出:</strong>"PINALSIGYAHRPI"
<strong>解释:</strong>
P     I    N
A   L S  I G
Y A   H R
P     I
</pre>

    <p><strong>示例 3:</strong></p>

    <pre><strong>输入:</strong>s = "A", numRows = 1
<strong>输出:</strong>"A"
</pre>

    <p>&nbsp;</p>

    <p><strong>提示:</strong></p>

    <ul>
        <li><code>1 &lt;= s.length &lt;= 1000</code></li>
        <li><code>s</code> 由英文字母(小写和大写)、<code>','</code> 和 <code>'.'</code> 组成</li>
        <li><code>1 &lt;= numRows &lt;= 1000</code></li>
    </ul>
</div>

## template

```python
import math
class Solution:
	def convert(self, s: str, numRows: int) -> str:
		n = len(s)
		N = numRows
		if n == 1 or N == 1:
			return s
		S = N-2
		C = 2*N-2
		R = int(math.floor(n/C))
		RS = n % (C)
		CE = n-R*C
		RR = 1 if (RS <= N) else 1+(RS-N)
		RX = R*(N-1) + RR
		output = []
		i = 0
		while i < N:
			j = 0
			k = (N-1-i)
			while j < RX:
				r = int(math.floor(j/(N-1)))
				rs = j % (N-1)
				offset = i if rs == 0 else N+rs-1
				index = r*C+offset
				if index < len(s):
					output.append(s[index])
				if i > 0 and i < N-1:
					r = int(math.floor(k/(N-1)))
					rs = k % (N-1)
					offset = i if rs == 0 else N+rs-1
					index = r*C+offset
					if index < len(s):
						output.append(s[index])
				j += (N-1)
				k += (N-1)
			i += 1
		return ''.join(output)
# %%
s = Solution()
print(s.convert('PAYPALISHIRING', 3))
```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```