6. ZigZag Conversion

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   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".

链接: http://leetcode.com/problems/zigzag-conversion/

题解:

根据例子列出n = 3和n = 4的情况，计算出列间距zigSize = 2 * numRows - 2，所以每行元素为 i + n * zigSize以及斜线的计算公式，当i为行数时，斜线上元素为i + n * zigSize - 2 * i。

Time Complexity - O(n)， Space Complexity - O(n)。

public class Solution {public String convert(String s, int numRows) {int zigSize = 2 * numRows - 2;if(s == null || s.length() == 0 || zigSize <= 0)return s;StringBuilder result = new StringBuilder();for(int i = 0; i < numRows; i ++){for(int j = i; j < s.length(); j += zigSize){result.append(s.charAt(j));    if(i != 0 && i != numRows - 1 && j + zigSize - 2 * i < s.length() )result.append(s.charAt(j + zigSize - 2 * i));}}return result.toString();}
}

二刷:

Java:

要注意 zigSize = 2 * numSize - 2, zag = j + zigSize - 2 * i

public class Solution {public String convert(String s, int numRows) {int zigSize = 2 * numRows - 2;  // calculate zigSizeif (s == null || s.length() == 0 || zigSize <= 0) {return s;}StringBuilder sb = new StringBuilder();for (int i = 0; i < numRows; i++) {for (int j = i; j < s.length(); j += zigSize) {sb.append(s.charAt(j));if (i != 0 && i != numRows - 1 && j + zigSize - 2 * i < s.length()) { // calculate zagsb.append(s.charAt(j + zigSize - 2 * i));}}}    return sb.toString();   }
}

Python:

class Solution(object):def convert(self, s, numRows):""":type s: str:type numRows: int:rtype: str"""zigSize = 2 * numRows - 2if s == None or zigSize <= 0:return sres = ''for i in range(numRows):for j in range(i, len(s), zigSize):res += s[j]zag = j + zigSize - 2 * iif i not in (0, numRows - 1) and zag < len(s):res += s[zag]return res

三刷:

其实就是看了一遍二刷的解而已。

Java:

public class Solution {public String convert(String s, int numRows) {int zigSize = 2 * numRows - 2; if (s == null || s.length() == 0 || zigSize <= 0) return s;StringBuilder sb = new StringBuilder();for (int i = 0; i < numRows; i++) {for (int j = i; j < s.length(); j += zigSize) {sb.append(s.charAt(j));if (i != 0 && i != numRows - 1 && j + zigSize - 2 * i < s.length()) { // calculate zagsb.append(s.charAt(j + zigSize - 2 * i));}}}    return sb.toString();   }
}

Reference:

http://www.cnblogs.com/springfor/p/3889414.html

转载于:https://www.cnblogs.com/yrbbest/p/4430333.html

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