LeetCode——006 ZigZag Conversion

Description

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

** Solution: **

注意输入的字母排序是按照倒着的N排序的，然后让你根据这个倒置的N的图形进行按行排序输出
按照半个N，即:一个竖和一个行的循环将数据输入到每一行中
比如输入：0，1，2，3，4，5，6，7，8，9，10；4行
其形状为
0 6
1 5 7
2 4 8 10
3 9
按照半个N，即0，1，2，3 与4，5 为循环，然后确定半个N的大小为size = 2 * numRows - 2 = 6
半个N的下竖为, 存入行[i%size]，上斜存入[size - i % size];

class Solution {
public:
    string convert(string s, int numRows) {
        if (s.length() < 2 || numRows < 2)return s;
        string res = "";
        vector<string>rows(numRows, "");
        int size = 2 * numRows - 2;//半个N的大小
        for (int i = 0; i < s.length(); ++i)
        {
            int id = i % size;
            if (id < numRows)//竖下
                rows[id] += s[i];
            else//斜上
                rows[size - id] += s[i];
        }
        for (auto str : rows)
            res += str;
        return res;
    }
};