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 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

【题意】

字符串是zigzag形状/拉链形状来排列的（从左往右横着看），现在按行来读，获得新的string

【解答】

可以找到位置之间的规律，比如第一行的两个元素之间相差2*numRows-2。需要考虑的情况是，一列就可以完全排完和只有一行，此时直接return。

时间复杂度O(n)，空间复杂度O(n)

【代码】

class Solution {
public:
    string convert(string s, int numRows) {
        int len = s.length();
        if(len <= numRows or numRows == 1) 
            return s;
        int diff = 2 * numRows - 2;
        string res = "";
        for(int row=0; row<numRows; row++){
            res += s[row];
            int nextPos = row + diff;
            while(1){
                if(nextPos - 2*row >= len)
                    break;
                if(row != 0 && row != (numRows - 1))
                    res += s[nextPos - 2*row];
                if(nextPos < len){
                    res += s[nextPos];
                    nextPos += diff;
                 }
                 else
                    break;
            }
        }
        return res;
    }
};