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
题解
class Solution { | |
public String convert(String s, int numRows) { | |
if (numRows == 1) return s; | |
StringBuilder result = new StringBuilder(); | |
for (int i = 0; i < numRows; i++){ | |
result.append(build(s, i, i * 2, (numRows - 1) * 2)); | |
} | |
return result.toString(); | |
} | |
private String build(String s, int start, int step, int temp) { | |
StringBuilder result = new StringBuilder(); | |
while (start < s.length()){ | |
result.append(s.charAt(start)); | |
if (step != temp) { | |
step = temp - step; | |
} | |
start = start + step; | |
} | |
return result.toString(); | |
} | |
} |
优化
/** | |
* t 为指定行数 * 2 - 2 | |
* 0 0+t 0+2t | |
* 1 t-1 1+t 2t-1 1+2t | |
* 2 t-2 2+t 2t-2 2+2t | |
* 3 3+t 3+2t | |
*/ | |
class Solution1 { | |
public String convert(String s, int numRows) { | |
if (numRows == 1) return s; | |
StringBuilder result = new StringBuilder(); | |
int temp = (numRows - 1) * 2; | |
for (int i = 0; i < numRows; i++){ | |
for (int j = 0; j < s.length() - i; j += temp){ | |
result.append(s.charAt(i + j)); | |
if (i > 0 && i < numRows - 1 && j + temp - i < s.length()){ | |
result.append(s.charAt(j + temp - i)); | |
} | |
} | |
} | |
return result.toString(); | |
} | |
} |
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/zigzag-conversion