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Commit e26e506

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feat: update solutions to lc problem: No.0006 (doocs#4478)
No.0006.Zigzag Conversion
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‎solution/0000-0099/0006.Zigzag Conversion/README.md‎

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@@ -78,9 +78,9 @@ P I
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### 方法一:模拟
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我们用一个二维数组 $g$ 来模拟 $Z$ 字形排列的过程,其中 $g[i][j]$ 表示第 $i$ 行第 $j$ 列的字符。初始时 $i=0,ドル另外我们定义一个方向变量 $k,ドル初始时 $k=-1,ドル表示向上走。
81+
我们用一个二维数组 $g$ 来模拟 Z 字形排列的过程,其中 $g[i][j]$ 表示第 $i$ 行第 $j$ 列的字符。初始时 $i = 0,ドル另外我们定义一个方向变量 $k,ドル初始时 $k = -1,ドル表示向上走。
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我们从左到右遍历字符串 $s,ドル每次遍历到一个字符 $c,ドル将其追加到 $g[i]$ 中,如果此时 $i=0$ 或者 $i=numRows-1,ドル说明当前字符位于 $Z$ 字形排列的拐点,我们将 $k$ 的值反转,即 $k=-k$。接下来,我们将 $i$ 的值更新为 $i+k,ドル即向上或向下移动一行。继续遍历下一个字符,直到遍历完字符串 $s,ドル我们返回 $g$ 中所有行拼接后的字符串即可。
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我们从左到右遍历字符串 $s,ドル每次遍历到一个字符 $c,ドル将其追加到 $g[i]$ 中。如果此时 $i = 0$ 或者 $i = \textit{numRows} - 1,ドル说明当前字符位于 Z 字形排列的拐点,我们将 $k$ 的值反转,即 $k = -k$。接下来,我们将 $i$ 的值更新为 $i + k,ドル即向上或向下移动一行。继续遍历下一个字符,直到遍历完字符串 $s,ドル我们返回 $g$ 中所有行拼接后的字符串即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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@@ -199,29 +199,24 @@ function convert(s: string, numRows: number): string {
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```rust
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impl Solution {
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pub fn convert(s: String, num_rows: i32) -> String {
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let num_rows = num_rows as usize;
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if num_rows == 1 {
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return s;
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}
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let mut ss = vec![String::new(); num_rows];
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let num_rows = num_rows as usize;
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let mut g = vec![String::new(); num_rows];
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let mut i = 0;
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let mut to_down = true;
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let mut k = -1;
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for c in s.chars() {
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ss[i].push(c);
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if to_down {
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i += 1;
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} else {
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i -= 1;
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}
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g[i].push(c);
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if i == 0 || i == num_rows - 1 {
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to_down = !to_down;
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k = -k;
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}
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i = (i as isize + k) as usize;
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}
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let mut res = String::new();
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for i in 0..num_rows {
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res += &ss[i];
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}
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res
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g.concat()
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}
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}
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```
@@ -323,220 +318,42 @@ char* convert(char* s, int numRows) {
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def convert(self, s: str, numRows: int) -> str:
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if numRows == 1:
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return s
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group = 2 * numRows - 2
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ans = []
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for i in range(1, numRows + 1):
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interval = group if i == numRows else 2 * numRows - 2 * i
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idx = i - 1
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while idx < len(s):
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ans.append(s[idx])
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idx += interval
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interval = group - interval
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if interval == 0:
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interval = group
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return ''.join(ans)
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```
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#### Java
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```java
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class Solution {
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public String convert(String s, int numRows) {
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if (numRows == 1) {
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return s;
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}
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StringBuilder ans = new StringBuilder();
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int group = 2 * numRows - 2;
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for (int i = 1; i <= numRows; i++) {
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int interval = i == numRows ? group : 2 * numRows - 2 * i;
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int idx = i - 1;
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while (idx < s.length()) {
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ans.append(s.charAt(idx));
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idx += interval;
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interval = group - interval;
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if (interval == 0) {
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interval = group;
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}
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}
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}
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return ans.toString();
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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string convert(string s, int numRows) {
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if (numRows == 1) return s;
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string ans;
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int group = 2 * numRows - 2;
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for (int i = 1; i <= numRows; ++i) {
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int interval = i == numRows ? group : 2 * numRows - 2 * i;
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int idx = i - 1;
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while (idx < s.length()) {
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ans.push_back(s[idx]);
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idx += interval;
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interval = group - interval;
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if (interval == 0) interval = group;
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}
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func convert(s string, numRows int) string {
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if numRows == 1 {
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return s
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}
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n := len(s)
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ans := make([]byte, n)
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step := 2*numRows - 2
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count := 0
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for i := 0; i < numRows; i++ {
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for j := 0; j+i < n; j += step {
421-
ans[count] = s[i+j]
422-
count++
423-
if i != 0 && i != numRows-1 && j+step-i < n {
424-
ans[count] = s[j+step-i]
425-
count++
426-
}
427-
}
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}
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return string(ans)
430-
}
431-
```
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#### TypeScript
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```ts
436-
function convert(s: string, numRows: number): string {
437-
if (numRows === 1) {
438-
return s;
439-
}
440-
const ss = new Array(numRows).fill('');
441-
let i = 0;
442-
let toDown = true;
443-
for (const c of s) {
444-
ss[i] += c;
445-
if (toDown) {
446-
i++;
447-
} else {
448-
i--;
449-
}
450-
if (i === 0 || i === numRows - 1) {
451-
toDown = !toDown;
452-
}
453-
}
454-
return ss.reduce((r, s) => r + s);
455-
}
456-
```
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#### Rust
459-
460-
```rust
461-
impl Solution {
462-
pub fn convert(s: String, num_rows: i32) -> String {
463-
let num_rows = num_rows as usize;
464-
let mut rows = vec![String::new(); num_rows];
465-
let iter = (0..num_rows).chain((1..num_rows - 1).rev()).cycle();
466-
iter.zip(s.chars()).for_each(|(i, c)| rows[i].push(c));
467-
rows.into_iter().collect()
468-
}
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}
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```
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#### JavaScript
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```js
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/**
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* @param {string} s
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* @param {number} numRows
478-
* @return {string}
479-
*/
480-
var convert = function (s, numRows) {
481-
if (numRows == 1) return s;
482-
const arr = new Array(numRows);
483-
for (let i = 0; i < numRows; i++) arr[i] = [];
484-
let mi = 0,
485-
isDown = true;
486-
for (const c of s) {
487-
arr[mi].push(c);
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if (mi >= numRows - 1) isDown = false;
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else if (mi <= 0) isDown = true;
491-
492-
if (isDown) mi++;
493-
else mi--;
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}
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let ans = [];
496-
for (const item of arr) {
497-
ans = ans.concat(item);
498-
}
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return ans.join('');
500-
};
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```
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#### PHP
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```php
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class Solution {
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/**
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* @param string $s
509-
* @param int $numRows
510-
* @return string
326+
* @param String $s
327+
* @param Integer $numRows
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* @return String
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*/
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function convert($s, $numRows) {
514-
if ($numRows == 1 || strlen($s) <= $numRows) {
331+
if ($numRows == 1) {
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return $s;
516333
}
517334
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$result = '';
519-
$cycleLength = 2 * $numRows - 2;
520-
$n = strlen($s);
335+
$g = array_fill(0, $numRows, '');
336+
$i = 0;
337+
$k = -1;
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for ($i = 0; $i < $numRows; $i++) {
523-
for ($j = 0; $j + $i < $n; $j += $cycleLength) {
524-
$result .= $s[$j + $i];
339+
$length = strlen($s);
340+
for ($j = 0; $j < $length; $j++) {
341+
$c = $s[$j];
342+
$g[$i] .= $c;
525343
526-
if ($i != 0 && $i != $numRows - 1 && $j + $cycleLength - $i < $n) {
527-
$result .= $s[$j + $cycleLength - $i];
528-
}
344+
if ($i == 0 || $i == $numRows - 1) {
345+
$k = -$k;
529346
}
530-
}
531347
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return $result;
348+
$i += $k;
349+
}
350+
return implode('', $g);
533351
}
534352
}
535-
``
353+
```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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```

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