6. ZigZag Conversion
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题目:
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 NA P L S I I GY I RAnd 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"
.题意:
给定按照Z字形放置的字符串以及Z字形放置的行数,将给定的字符串按照逐行读取的方式读取之后返回。
思路:
找数学规律如下:
所以,对于每一层垂直元素的坐标(i; j) = (j + 1) * n + i;对于每两层垂直元素之间的插入元素(斜对角元素),(i; j) = (j + 1) * n - i。
代码:C++版:16ms
class Solution {public: string convert(string s, int numRows) { if (numRows <= 1 || s.size() <= 1) return s; string result; for (int i = 0; i < numRows; i++) { for (int j = 0, index = i; index < s.size(); j++, index = (2*numRows - 2)*j+i) { result.append(1, s[index]); //垂直元素 if (i == 0 || i == numRows - 1) continue; //斜对角元素 if (index + (numRows - i - 1) * 2 < s.size()) result.append(1, s[index + (numRows - i - 1) * 2]); } } return result; }};
代码:Java版:13ms
public class Solution { public String convert(String s, int numRows) { char[] c = s.toCharArray(); int len = c.length; StringBuffer[] sb = new StringBuffer[numRows]; for (int i=0; i<sb.length; i++) { sb[i] = new StringBuffer(); } int i = 0; while (i < len) { for (int idx=0; idx<numRows && i<len; idx++) sb[idx].append(c[i++]); for (int idx=numRows-2; idx>=1 && i<len; idx--) sb[idx].append(c[i++]); } for (int idx=1; idx<sb.length; idx++) sb[0].append(sb[idx]); return sb[0].toString(); }}
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