LeetCode-ZigZag Convert

题目描述

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

思路

本题重点在于使用字符串vector来保存ZigZag之后的字符数组，并将二维转换等价为一维，因此只需要控制行的移动即可，此外，还需注意每次移动的方向，定义bool变量来控制。

程序代码

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows == 1)
            return s;
        int len = s.length();
        //使用vector存储二维数组，vector的构造函数是容量大小
        vector<string> rows(min(len,numRows));
        int currRow = 0;
        bool goUp = false;
        for(int i=0;i<len;i++){
            rows[currRow] += s.at(i);
            // printf("row:%d,char:%c\n",currRow,s.at(i) );
            //判断当前位置是否向下还是向上
            if(currRow == numRows - 1){
                goUp = true;
            }else if(currRow == 0){
                goUp = false;
            }
            if(goUp){
                currRow --;
            }else{
                currRow ++;
            }

        }
        string result ;
        for(int i=0;i<rows.size();i++){
            result += rows[i];
        }
        return result;
    }
};