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
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)
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();
}
}
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;
}
};
using System.Collections.Generic;
using System.Linq;
public class Solution {
public string Convert(string s, int numRows) {
if (numRows == 1) return s;
if (numRows > s.Length) numRows = s.Length;
var rows = new List<char>[numRows];
var i = 0;
var j = 0;
var down = true;
while (i < s.Length)
{
if (rows[j] == null)
{
rows[j] = new List<char>();
}
rows[j].Add(s[i]);
j = j + (down ? 1 : -1);
if (j == numRows || j < 0)
{
down = !down;
j = j + (down ? 2 : -2);
}
++i;
}
return new string(rows.SelectMany(row => row).ToArray());
}
}
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)
}
/**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function (s, numRows) {
if (numRows == 1) return s;
let 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 (let item of arr) {
ans = ans.concat(item);
}
return ans.join('');
};
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);
}
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
}
}
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()
}
}