LeetCode 6. ZigZag Conversion(java)

xiaoxiao2021-02-28  2

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”.

思路:通过观察这个N型的图案的组成,我们发现可以通过从上到下(0 ~ numRows - 1),再从下到上(numRows - 1, 1),以此为反复,就可以得到想要的排列。所以解决的方法就是创建numRows个StringBuilder,每个StringBuilder都可以储存每行的字母,通过N型的排列来将字母一个个加进对应的行里。

public String convert(String s, int numRows) { if (s.length() == 0 || numRows <= 1) return s; int length = s.length(), index = 0; StringBuilder[] sb = new StringBuilder[numRows]; for (int i = 0; i < numRows; i++) { sb[i] = new StringBuilder(); } while (index < length) { for (int i = 0; i < numRows && index < length; i++) { System.out.println(sb[i]); sb[i].append(s.charAt(index)); index++; } for (int j = numRows - 2; j >= 1 && index < length; j--) { sb[j].append(s.charAt(index)); index++; } } StringBuilder res = new StringBuilder(); for (int i = 0; i < numRows; i++) { res.append(sb[i]); } return res.toString(); }
转载请注明原文地址: https://www.6miu.com/read-1600132.html

最新回复(0)