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LeetCode:ZigZag Conversion

2024-07-05 08:07:00   225人阅读

题目链接

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".

题目的意思是把字符串上下上下走之字行,然后按行输出,比如包含数字0~22的字符串, 给定行数为5,走之字形如下:

image

现在要按行输出字符,即:0 8 16 1 7 9 15 17 2…….

如果把以上的数字字符看做是字符在原数组的下标, 给定行数为n = 5, 可以发现以下规律:

(1)第一行和最后一行下标间隔都是interval = n*2-2 = 8 ;                                                   本文地址

(2)中间行的间隔是周期性的,第i行的间隔是: interval–2*i,  2*i,  interval–2*i, 2*i, interval–2*i, 2*i, …

 

代码如下，时间复杂度为O（n），n是字符串的长度：

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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)//处理第一行
            res[k++] = s[j];
        for(int i = 1; i < nRows-1; i++)//处理中间行
        {
            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)//处理最后一行
            res[k++] = s[j];
        return res;
    }
};

 

【版权声明】转载请注明出处：http://www.cnblogs.com/TenosDoIt/p/3738693.html


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