[LeetCode] ZigZag Conversion

最新推荐文章于 2021-11-05 22:58:08 发布

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最新推荐文章于 2021-11-05 22:58:08 发布

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原文链接：http://www.cnblogs.com/changchengxiao/p/3594839.html

本文介绍了一种将字符串以Z形方式分布在多行上的算法，并实现了一个C++函数用于完成这种转换。例如，字符串PAYPALISHIRING在3行上按Z形排列后读取为PAHNAPLSIIGYIR。代码中详细展示了如何通过使用多个临时字符串来模拟这一过程。

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".

Solution:

class Solution {
public:
    string convert(string s, int nRows) {
        string *tmpS = new string[nRows + 1];
        string ans;
        if(nRows < 2) return s;
        
        bool down = true;
        int row = 0;
        
        for(int i = 0;i < s.length();i++)
        {
            if(row == -1)
            {
                row = 1;
                down = true;
                tmpS[row] += s[i];
                row++;
            }
            else if(row == nRows)
            {
                row = nRows - 2;
                down = false;
                tmpS[row] += s[i];
                row--;
            }
            else
            {
                tmpS[row] += s[i];
                if(down) row++;
                else row--;
            }
        }
        
        for(int i = 0;i < nRows;i++)
            ans += tmpS[i];
        return ans;
    }
};