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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”.

题目大意

• 1.之字型摆字符串然后横着输出重组了的字符串。

解题思路

• 1.字符串最后是这么摆的，每行都由主元素和插入元素组成，主元素中间插入一个元素，每层的主元素间隔为2*(n-1),插入的元素与它前面一个元素的间隔为2×(n-1)-2*i。

AC代码

#include<iostream>
#include<vector>
#include<cstdio>
using namespace std;

class Solution {
public:
    string convert(string s, int numRows) {
        if (s.empty()) {
            return s;
        }
        int length = s.length();
        if (length<=numRows||numRows==1) {
            return s;
        }
        string sb = "";
        //主元素之间的间隔
        int step = 2*(numRows-1);
        int count = 0;
        for (int i = 0; i < numRows; ++i) {
            //主元素之间插入元素与它前一个主元素的间隔
            int interval = step - 2*i;
            for (int j = i; j < length; j+=step) {
                sb = sb + s[j];
                //满足条件则插入
                if (interval>0&&interval<step&&j+interval<length) {
                    sb = sb+s[j+interval];
                }
            }
        }
        return sb;
    }
};

int main(int argc, char *argv[]) {
    string aaa = "PAYPALISHIRING";
    Solution solution;
    aaa = solution.convert(aaa,3);
    cout << aaa << endl;
}