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LeetCode: ZigZag Conversion 解题报告
ZigZag Conversion
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".
SOLUTION 1:
使用以下算法会比较简单。
两个规律:
1 两个zigzag之间间距为2*nRows-2
2 每个zigzag中间(在j和j+interval之间)位置为j+interval-2*i
注意:当Rows = 1时,此方法不适用,因为size = 0,会造成死循环。所以Rows = 1时,需要独立处理。
引自:http://blog.csdn.net/fightforyourdream/article/details/16881517
1 public class Solution { 2 public String convert(String s, int nRows) { 3 if (s == null) { 4 return null; 5 } 6 7 // 第一个小部分的大小 8 int size = 2 * nRows - 2; 9 10 // 当行数为1的时候,不需要折叠。11 if (nRows <= 1) {12 return s;13 }14 15 StringBuilder ret = new StringBuilder();16 17 int len = s.length();18 for (int i = 0; i < nRows; i++) {19 // j代表第几个BLOCK20 for (int j = i; j < len; j += size) {21 ret.append(s.charAt(j));22 23 // 即不是第一行,也不是最后一行,还需要加上中间的节点24 int mid = j + size - i * 2;25 if (i != 0 && i != nRows - 1 && mid < len) {26 char c = s.charAt(mid);27 ret.append(c);28 }29 }30 }31 32 return ret.toString();33 }34 }
请至主页君的GIT HUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/Convert.java
LeetCode: ZigZag Conversion 解题报告