[leetcode]6. ZigZag Conversion字符串Z形排列

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 s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"

Explanation:

P     I      N
A   L S   I  G
Y A   H  R
P     I

 

Solution1: 

Step1: new StringBuilder[]. For each row, allocate a new StringBuilder to save characters in such row. 

         Pay attention! StringBuilder is treated like variable-length arrays. So StringBuilder[] is like array of array

Step2: we can observe that for 1st and last row,  chars will be added into sb[0] and sb[numRows-1], seperately.

                                       for sloping slide,  chars will be added in sb[numRows -2 ] ... sb[1]

Step3: convert each row's StringBuilder into result String

 

code:

/*
Time Complexity: O(n)
Space Complexity: O(n)
*/

class Solution {
   public String convert(String s, int numRows) {
        char[] c = s.toCharArray();
        int len = c.length;
        StringBuilder[] sb = new StringBuilder[numRows];
        // 要按row来进行遍历，每一个row先allocate一个StringBuilder
        for (int i = 0; i < sb.length; i++) {
            sb[i] = new StringBuilder();
        }

        int idx = 0; //用来扫String s
        while (idx < len) {
            for (int i = 0; i < numRows && idx < len; i++) {
                sb[i].append(c[idx++]);
            }
            // sloping side
            for (int i = numRows - 2; i >= 1 && idx < len; i--) {
                sb[i].append(c[idx++]);
            }

        }
        //从sb[0]开始，将sb[1], sb[2], sb[3]... append到一个StringBuilder
        for (int i = 1; i < sb.length; i++) {
            sb[0].append(sb[i]);
        }
        return sb[0].toString();
    }
}