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  • ZigZag Conversion

    题目描述:

    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 text, int nRows);

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


      理解题目意思很关键。

      nRows = 3,字母排列如下:

      P          A

      A    P    L

      Y          I           (按Z字形排列)

      nRows = 4,字母排列如下:

      P               I

      A         L    S

      Y    A         H 

      P               I      (按Z字形排列)

      


      代码很简单,自己写了一个

    solution1:

    string convert(string s, int nRows) {
        int len = s.size();
        if(len <= 2 || nRows <=1)
            return s;
        if(len <= nRows)
            return s;
        vector<string> res(nRows);
        int num = 2 * (nRows-1);
        int *loop = new int[num];
        int i,j;
        for (i = 0;i < nRows;++i)
        {
            loop[i] = i;
        }
        for (j = nRows-2;j > 0;--j)
        {
            loop[i++] = j;
        }
        int count = len / num;
        i = 0;
        for (j = 0;j < count;++j)
        {
            for (int k = 0;k < num;++k)
            {
                res[loop[k]] += s[i++];
            }
        }
        int k = 0;
        for (;i < len;++i)
        {
            res[loop[k++]] += s[i];
        }
        for (i = 1;i < nRows;++i)
        {
            res[0] += res[i];
        }
        delete [] loop;
        return res[0];
    }

      看上去太过粗糙,下面的更精简
    solution2:

    string convert(string s, int nRows) {
        if(nRows < 2)
            return s;
        int len = s.size();
        vector<string> res(nRows);
        int i = 0;
        while (i < len)
        {
            for (int idx = 0;idx < nRows && i < len;idx++)   // vertically down
                res[idx] += s[i++];
            for (int idx = nRows-2;idx >=1 && i < len;idx--) // obliquely up
                res[idx] += s[i++];
        }
        for (int idx = 1;idx < nRows;idx++)
            res[0] += res[idx];
        return res[0];
    }

    来源:https://oj.leetcode.com/discuss/10493/easy-to-understand-java-solution

      

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  • 原文地址:https://www.cnblogs.com/gattaca/p/4174648.html
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