ZigZag Conversion
Description
The string "PAYPALISHIRING"
is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string text, int nRows);
convert("PAYPALISHIRING", 3)
should return "PAHNAPLSIIGYIR"
.
Tags: String
思路 0
题意是让你把字符串按波浪形排好,然后返回横向读取的字符串。
听不懂的话,看下面的表示应该就明白了:
0 2n-2 4n-4
1 2n-3 2n-1 4n-5 4n-5
2 2n-4 2n 4n-6 .
. . . . .
. n+1 . 3n-1 .
n-2 n 3n-4 3n-2 5n-6
n-1 3n-3 5n-5
那么我们可以根据上面找规律,可以看到波峰和波谷是单顶点的,它们周期是 2 * (n - 1)
,单独处理即可;中间的部分每个周期会出现两次,规律很好找,留给读者自己想象,不懂的可以结合以下代码。
class Solution {
public String convert(String s, int numRows) {
if (numRows <= 1) return s;
int len = s.length();
char[] chars = s.toCharArray();
int cycle = 2 * (numRows - 1);
StringBuilder sb = new StringBuilder();
for (int j = 0; j < len; j += cycle) {
sb.append(chars[j]);
}
for (int i = 1; i < numRows - 1; i++) {
int step = 2 * i;
for (int j = i; j < len; j += step) {
sb.append(chars[j]);
step = cycle - step;
}
}
for (int j = numRows - 1; j < len; j += cycle) {
sb.append(chars[j]);
}
return sb.toString();
}
}
思路 1
另外一种思路就是开辟相应行数的 StringBuilder
对象,然后模拟波浪生成的样子分别插入到相应的 StringBuilder
对象,比较直白简单,具体代码如下。
class Solution {
public String convert(String s, int numRows) {
if (numRows <= 1) return s;
int len = s.length();
char[] chars = s.toCharArray();
StringBuilder[] sbs = new StringBuilder[numRows];
for (int i = 0; i < numRows; i++) {
sbs[i] = new StringBuilder();
}
int i = 0;
while (i < len) {
for (int j = 0; j < numRows && i < len; ++j) {
sbs[j].append(chars[i++]);
}
for (int j = numRows - 2; j >= 1 && i < len; --j) {
sbs[j].append(chars[i++]);
}
}
for (i = 1; i < numRows; i++) {
sbs[0].append(sbs[i]);
}
return sbs[0].toString();
}
}
结语
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