6. ZigZag Conversion（Z字形输出数组）

官网

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;}