LeetCode:ZigZag Conversion

題目連接

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".

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題目的意思是把字符串上下上下走之字形狀,而後按行輸出,好比包含數字0~22的字符串, 給定行數爲5,走之字形以下:

如今要按行輸出字符,即:0 8 16 1 7 9 15 17 2…….

若是把以上的數字字符看作是字符在原數組的下標, 給定行數爲n = 5, 能夠發現如下規律:

(1)第一行和最後一行下標間隔都是interval = n*2-2 = 8 ;                                                   本文地址

(2)中間行的間隔是週期性的,第i行的間隔是: interval–2*i,  2*i,  interval–2*i, 2*i, interval–2*i, 2*i, …

 

代碼以下，時間複雜度爲O（n），n是字符串的長度：

class Solution {
public:
    string convert(string s, int nRows) {
        if(nRows == 1)return s;
        int len = s.size(), k = 0, interval = (nRows<<1)-2;
        string res(len, ' ');
        for(int j = 0; j < len ; j += interval)//處理第一行
            res[k++] = s[j];
        for(int i = 1; i < nRows-1; i++)//處理中間行
        {
            int inter = (i<<1);
            for(int j = i; j < len; j += inter)
            {
                res[k++] = s[j];
                inter = interval - inter;
            }
        }
        for(int j = nRows-1; j < len ; j += interval)//處理最後一行
            res[k++] = s[j];
        return res;
    }
};

 

【版權聲明】轉載請註明出處：http://www.cnblogs.com/TenosDoIt/p/3738693.html