ZigZag Conversion

Question

• leetcode: ZigZag Conversion| LeetCode OJ

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".

題解

可以發(fā)現(xiàn)首行和末行的重復(fù)周期都是 2 * nRows - 2，對于首行和末行之間的行，還會額外重復(fù)一次，重復(fù)的這一次距離本周期起始字符的距離是 2 * nRows - 2 - 2 * i 。

C++

class Solution {
public:
    string convert(string s, int nRows) {
        if (nRows <= 1 || s.length() == 0)  //nRows=1 按下面會死循環(huán)
           return s;  
        string ret = "";
        const int len = s.size();
        int lag = 2*nRows - 2; //循環(huán)周期  
        for (int i = 0; i < nRows; i++) {  
            for (int j = i; j < len; j += lag) {  
                ret += s[j];    
                if (i > 0 && i < nRows-1) {  //非首行和末行時
                    int t = j + lag - 2*i;  
                    if (t < len) {  
                        ret += s[t];  
                    }  
                }  
            }  
        }  
        return ret;
   
    }
};

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