6. ZigZag Conversion

作者: 七八音 | 来源:发表于2018-12-17 22:32 被阅读1次

    题目

    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

    设置numRows个数组,按照顺序依次添加到对应数组中;同时设置步长,用来修改对应的行号:
    row == 0: step=1(依次往下,row++);row == numRows-1: step=-1(到最后一行,依次往上,row--行号递减)
    row += step;
    最后将各行结果依次添加到新字符串中.
    将各行结果存储在对应字符串中,不用计算转换后的下标,然后拼接最终结果了.
    pros & cons: 容易理解; 浪费存储空间

    class Solution {
    public:
        string convert(string s, int numRows) {
            if(numRows < 2) return s;
            int len = s.size();
            string* str = new string[numRows];
            int row = 0, step = 1;
            for(int i=0;i<len;i++){
                str[row].push_back(s[i]);
                if(row == 0) step = 1;
                else if(row == numRows-1) step = -1;
                row += step;
            }
            string res;
            for(int i=0;i<numRows;i++)
                res.append(str[i]);
            delete[] str;
            
            return res;
        }
    };
    

    相关文章

      网友评论

        本文标题:6. ZigZag Conversion

        本文链接:https://www.haomeiwen.com/subject/lsyhkqtx.html