Leecode Week5: ZigZag Conversion

Week5: ZigZag Conversion

Difficulty:Medium

1.Question

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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

2.Algorithm

1.首先根據輸入的numRows分爲numRows個字符串——string row[numRows]，該數組每i個元素代表着第i行的數據

2.接下來判斷s中的每個字符屬於哪一行，判斷的依據如下：
divisor = 2 * (numRows - 1)
temp = 字符所在的索引 / divisor
若temp <= numRows，則temp即爲字符所屬行
若temp >= numRows，則divisor-numRows即爲字符所屬行

3.將判斷好的字符加入對應的row數組中

4.最後按順序將四個字符串加起來便爲輸出的結果

3.Solution

class Solution {
public:
    string convert(string s, int numRows) {
        string row[numRows];
        string result = "";
        for (int i = 0; i < numRows; i++) {
            row[i] = "";
        }
        int divisor = 1;
        if (numRows > 1) {
            divisor = 2 * (numRows-1);
        }
        for (int i = 0; i < s.length(); i++) {
            int temp = i % divisor;
            if(temp > numRows - 1) {
                temp = divisor - temp;
            }
            row[temp] += s[i];
        }
        for(int i = 0; i < numRows; i++) {
            result += row[i];
        }
        return result;
    }
};