[LeetCode]#6 ZigZag Conversion

GitHub : https://github.com/MummyDing/LeetCode

Source :  https://leetcode.com/problems/zigzag-conversion/

Description :

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" .

题解: 总算有个1A了,不过也debug好久. 题目的意思是给定一个字符串,你需要以蛇形排列成锯齿图形.其实这就是一道找规律的题. 我假设字符串长度为len,需要排列成R行的锯齿图形,那么就有:

------------这里每个数字表示字符对应的下标

根据这个就可以找出规律

1. 第一行和最后一行: 横向递增公差为 2R-2

2. 中间的行:奇数列(列数从0开始标号)与后一行相加之和 = (col*2+2)*(R - 1)

3.  中间的行:偶数列(列数从0开始标号)与后一行相加之和 = (col+1)*(numRows - 1)*2

根据这个几个规律就很容易写代码了.(我谢了一面A4纸...囧...)

/*
Problem: ZigZag Conversion
Description: https://leetcode.com/problems/zigzag-conversion/
Author: MummyDing
Date : 2015-12-30
Run Time: 20 ms
*/
#include <iostream>

using namespace std;

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows <= 1) return s;
        string res = "";
        int len = s.size();
        for(int i=0 ; i<numRows ; i++){
            int col = 0 ;
            for( int j=i ; j<len ; ){
                res += s[j];
                if( i== 0 || i == numRows-1 ){
                    j += (2*numRows - 2);
                }else if(col % 2 == 0){
                    j = ((col*2+2)*(numRows - 1) -j);
                }else {
                    j = ((col+1)*(numRows - 1)*2 -j);
                }
                col++;
            }
        }
        return res;
    }
};

int main()
{
    //Solution s;
    //cout<<s.convert("1234567890abcd",6)<<endl;
    return 0;
}

【转载请注明出处】

Author: MummyDing

出处:http://blog.csdn.net/u012560612/article/category/6047025