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LeetCode #6 ZigZag Conversion (E)

2015-10-01 23:51 246 查看

[Problem]

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"

.

[Analysis]

这题比较简单，大概有两类思路：一是找到规律按一定的间隔直接将字符存入buffer，二是分开每行分别建立字符串最后将所有字符串连接起来。前者规律要稍微复杂一点，但内存方面稍微优化。一开始我采用的是将例子写在纸上找规律的暴力方法，只要将规律列清楚，一样可以求解。

[Solution]

public class Solution {
public String convert(String s, int nRows) {
if (s.length() <= nRows || nRows == 1) {
return s;
}

StringBuffer buf = new StringBuffer("");

for (int i = 0; i < nRows; i++) {
int interval1 = (nRows - i - 1) * 2;
int interval2 = i * 2;
if (interval2 == 0 || interval1 == interval2) {
for (int j = i; j < s.length(); j += interval1) {
buf.append(s.charAt(j));
}
} else if (interval1 == 0) {
for (int j = i; j < s.length(); j += interval2) {
buf.append(s.charAt(j));
}
} else {
buf.append(s.charAt(i));
for (int j = i + interval1; j < s.length(); j += interval1) {
buf.append(s.charAt(j));
j += interval2;
if (j < s.length()) {
buf.append(s.charAt(j));
}
}
}
}

return buf.toString();
}
}

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