題目描述

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);

通俗一點來說，就是將給定字串以N字形輸出之後再去掉空格回車按行順序得到一個全新的字串。
來看幾個例子：

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

求解演算法

一開始沒有注意到要去掉空格，發現之後全都註釋掉對應部分就過了。思路很簡單，就是發現規律而已。

class Solution {
public:
    string convert(string s, int numRows) {
    	if (numRows == 1) {
    		return s;
		}
        int L = s.length();
        int a = L/(2*numRows-2), b = L%(2*numRows-2);
        int numCols = a*(numRows-1)+1+b/numRows*(b%numRows);
        string ss = "";
        int index =
 
 0;
        for (int i = 0; i < numRows; i++) {
            for (int j = 0; j < numCols; j++) {
                a = j%(numRows-1);
                if (a > 0 && a+i != numRows-1) {
                    //ss += " ";
                }
                else {
                	b = j/(numRows-1);
                	index = b*(2*numRows-2)+(a==0?i:numRows+a-1);
                	if (index < s.length()) {
                		ss += s[index] ;
					}
					/*else {
						ss += ' ';
					}*/
                }
            }
            //ss += '\n';
        }
        return ss;
    }
};

其實還有一種思路就是，先獲得每一列對應的字串，然後按行拼接起來，這樣很容易實現，但是複雜度就上去了。