题目描述
将一个给定字符串 s
根据给定的行数 numRows
,以从上往下、从左到右进行 Z 字形排列。
比如输入字符串为 "PAYPALISHIRING"
行数为 3
时,排列如下:
P A H N
A P L S I I G
Y I R
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:"PAHNAPLSIIGYIR"
。
请你实现这个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);
示例 1:
输入:s = "PAYPALISHIRING", numRows = 3
输出:"PAHNAPLSIIGYIR"
示例 2:
输入:s = "PAYPALISHIRING", numRows = 4
输出:"PINALSIGYAHRPI"
解释:
P I N
A L S I G
Y A H R
P I
示例 3:
输入:s = "A", numRows = 1
输出:"A"
提示:
1 <= s.length <= 1000
s
由英文字母(小写和大写)、','
和 '.'
组成
1 <= numRows <= 1000
解法
方法一:模拟
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 $i=0$,另外我们定义一个方向变量 $k$,初始时 $k=-1$,表示向上走。
我们从左到右遍历字符串 $s$,每次遍历到一个字符 $c$,将其追加到 $g[i]$ 中,如果此时 $i=0$ 或者 $i=numRows-1$,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 $i+k$,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 $s$,我们返回 $g$ 中所有行拼接后的字符串即可。
时间复杂度 $O(n)$,空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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12 | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
g = [[] for _ in range(numRows)]
i, k = 0, -1
for c in s:
g[i].append(c)
if i == 0 or i == numRows - 1:
k = -k
i += k
return ''.join(chain(*g))
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18 | class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) {
return s;
}
StringBuilder[] g = new StringBuilder[numRows];
Arrays.setAll(g, k -> new StringBuilder());
int i = 0, k = -1;
for (char c : s.toCharArray()) {
g[i].append(c);
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
return String.join("", g);
}
}
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22 | class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) {
return s;
}
vector<string> g(numRows);
int i = 0, k = -1;
for (char c : s) {
g[i] += c;
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
string ans;
for (auto& t : g) {
ans += t;
}
return ans;
}
};
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15 | func convert(s string, numRows int) string {
if numRows == 1 {
return s
}
g := make([][]byte, numRows)
i, k := 0, -1
for _, c := range s {
g[i] = append(g[i], byte(c))
if i == 0 || i == numRows-1 {
k = -k
}
i += k
}
return string(bytes.Join(g, nil))
}
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16 | function convert(s: string, numRows: number): string {
if (numRows === 1) {
return s;
}
const g: string[][] = new Array(numRows).fill(0).map(() => []);
let i = 0;
let k = -1;
for (const c of s) {
g[i].push(c);
if (i === numRows - 1 || i === 0) {
k = -k;
}
i += k;
}
return g.flat().join('');
}
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27 | impl Solution {
pub fn convert(s: String, num_rows: i32) -> String {
let num_rows = num_rows as usize;
if num_rows == 1 {
return s;
}
let mut ss = vec![String::new(); num_rows];
let mut i = 0;
let mut to_down = true;
for c in s.chars() {
ss[i].push(c);
if to_down {
i += 1;
} else {
i -= 1;
}
if i == 0 || i == num_rows - 1 {
to_down = !to_down;
}
}
let mut res = String::new();
for i in 0..num_rows {
res += &ss[i];
}
res
}
}
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21 | /**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function (s, numRows) {
if (numRows === 1) {
return s;
}
const g = new Array(numRows).fill(_).map(() => []);
let i = 0;
let k = -1;
for (const c of s) {
g[i].push(c);
if (i === 0 || i === numRows - 1) {
k = -k;
}
i += k;
}
return g.flat().join('');
};
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25 | public class Solution {
public string Convert(string s, int numRows) {
if (numRows == 1) {
return s;
}
int n = s.Length;
StringBuilder[] g = new StringBuilder[numRows];
for (int j = 0; j < numRows; ++j) {
g[j] = new StringBuilder();
}
int i = 0, k = -1;
foreach (char c in s.ToCharArray()) {
g[i].Append(c);
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
StringBuilder ans = new StringBuilder();
foreach (StringBuilder t in g) {
ans.Append(t);
}
return ans.ToString();
}
}
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方法二
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16 | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
group = 2 * numRows - 2
ans = []
for i in range(1, numRows + 1):
interval = group if i == numRows else 2 * numRows - 2 * i
idx = i - 1
while idx < len(s):
ans.append(s[idx])
idx += interval
interval = group - interval
if interval == 0:
interval = group
return ''.join(ans)
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22 | class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) {
return s;
}
StringBuilder ans = new StringBuilder();
int group = 2 * numRows - 2;
for (int i = 1; i <= numRows; i++) {
int interval = i == numRows ? group : 2 * numRows - 2 * i;
int idx = i - 1;
while (idx < s.length()) {
ans.append(s.charAt(idx));
idx += interval;
interval = group - interval;
if (interval == 0) {
interval = group;
}
}
}
return ans.toString();
}
}
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19 | class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) return s;
string ans;
int group = 2 * numRows - 2;
for (int i = 1; i <= numRows; ++i) {
int interval = i == numRows ? group : 2 * numRows - 2 * i;
int idx = i - 1;
while (idx < s.length()) {
ans.push_back(s[idx]);
idx += interval;
interval = group - interval;
if (interval == 0) interval = group;
}
}
return ans;
}
};
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20 | func convert(s string, numRows int) string {
if numRows == 1 {
return s
}
n := len(s)
ans := make([]byte, n)
step := 2*numRows - 2
count := 0
for i := 0; i < numRows; i++ {
for j := 0; j+i < n; j += step {
ans[count] = s[i+j]
count++
if i != 0 && i != numRows-1 && j+step-i < n {
ans[count] = s[j+step-i]
count++
}
}
}
return string(ans)
}
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20 | function convert(s: string, numRows: number): string {
if (numRows === 1) {
return s;
}
const ss = new Array(numRows).fill('');
let i = 0;
let toDown = true;
for (const c of s) {
ss[i] += c;
if (toDown) {
i++;
} else {
i--;
}
if (i === 0 || i === numRows - 1) {
toDown = !toDown;
}
}
return ss.reduce((r, s) => r + s);
}
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| impl Solution {
pub fn convert(s: String, num_rows: i32) -> String {
let num_rows = num_rows as usize;
let mut rows = vec![String::new(); num_rows];
let iter = (0..num_rows).chain((1..num_rows - 1).rev()).cycle();
iter.zip(s.chars()).for_each(|(i, c)| rows[i].push(c));
rows.into_iter().collect()
}
}
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26 | /**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function (s, numRows) {
if (numRows == 1) return s;
const arr = new Array(numRows);
for (let i = 0; i < numRows; i++) arr[i] = [];
let mi = 0,
isDown = true;
for (const c of s) {
arr[mi].push(c);
if (mi >= numRows - 1) isDown = false;
else if (mi <= 0) isDown = true;
if (isDown) mi++;
else mi--;
}
let ans = [];
for (const item of arr) {
ans = ans.concat(item);
}
return ans.join('');
};
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29 | class Solution {
/**
* @param string $s
* @param int $numRows
* @return string
*/
function convert($s, $numRows) {
if ($numRows == 1 || strlen($s) <= $numRows) {
return $s;
}
$result = '';
$cycleLength = 2 * $numRows - 2;
$n = strlen($s);
for ($i = 0; $i < $numRows; $i++) {
for ($j = 0; $j + $i < $n; $j += $cycleLength) {
$result .= $s[$j + $i];
if ($i != 0 && $i != $numRows - 1 && $j + $cycleLength - $i < $n) {
$result .= $s[$j + $cycleLength - $i];
}
}
}
return $result;
}
}
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