LeetCode6-ZigZag Conversion(C++)

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

AC代碼

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows <= 1) {
            return s;
        }

        vector<string> pattern(numRows, "");
        string res = "";
        // down is a flag which determine the direction
        int down = 0;
        // row stands for which row it is now
        int row = 0;
        for(int i=0; i<s.size(); i++) {
            pattern[row].push_back(s[i]);
            if(row == 0) {
                down = 1;  // Move downside
            } else if(row == numRows - 1) {
                down = -1;  // Move upside
            }
            row += down;
        }

        for(int i=0; i<numRows; i++) {
            res += pattern[i];
        }

        return res;
    }
};

測試代碼

int main() {
    Solution s;

    string s1 = "PAYPALISHIRING";
    string res_s1 = s.convert(s1, 3);
    cout << res_s1 << endl;

    string s2 = "PAYPALISHIRING";
    string res_s2 = s.convert(s2, 4);
    cout << res_s2 << endl;
}

總結

使用犧牲空間複雜度的方式換取時間複雜度，創建一個vector，來存儲每一行的字符串。設置一個flag，決定index移動的方向。