ZigZag Conversion

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   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".

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题目的意思是把字符串上下上下走之字形状,然后按行输出,比如包含数字0~22的字符串, 给定行数为5,走之字形如下:

现在要按行输出字符,即:0 8 16 1 7 9 15 17 2…….

如果把以上的数字字符看做是字符在原数组的下标, 给定行数为n = 5, 可以发现以下规律:

(1)第一行和最后一行下标间隔都是interval = n*2-2 = 8 ;                                                   本文地址

(2)中间行的间隔是周期性的,第i行的间隔是: interval–2*i,  2*i,  interval–2*i, 2*i, interval–2*i, 2*i, …

 

代码如下，时间复杂度为O（n），n是字符串的长度：

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    string convert(string s, int numRows) {
        if(s == "")
            return "";
        if(numRows==1||s.length()<=numRows)
            return s;
        string result;
        int interval = numRows*2 -2;
        unsigned long len = s.length();
        int tmp = 0;
        bool flag = true;
        int x = 0;
        for(int i =0;i<numRows;i++){
            tmp = i;
            result.push_back(s[tmp]);
            while (tmp < len -1 ){
                x = tmp;
                if(flag){
                    tmp = tmp + interval - 2*i;
                    flag = false;
                }else{
                    tmp = tmp + 2*i;
                    flag = true;
                }
                if(x!=tmp && tmp < len)
                    result.push_back(s[tmp]);

            }
            flag = true;
        }
        return result;
    }
};
int main() {
    Solution s;
    cout<<s.convert("A",2);
    return 0;
}