64. ZigZag Conversion

ZigZag Conversion

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".

思路：

方法1： 刚开始正着方三个，然后反着放两个，正着放两个。直到结束。（优）

class Solution {
public:
    string convert(string s, int nRows) {
        if(nRows <= 1 || s == "") return s;
        int len = s.length();
		string s2[nRows];
        int k = 0, k2 = 0;
        while(k2 < len){
            while(k < nRows && k2 < len) s2[k++].push_back(s[k2++]);
			k--;
            while(k > 0 && k2 < len) s2[--k].push_back(s[k2++]);
			k++;
        }
        string s3;
        for(int i = 0; i < nRows; ++i) {
            s3 += s2[i];
        }
        return s3;
    }
};

 方法二： 类似方法一，不过存下每个位置的字符应该放的地方。然后依次读入。

class Solution {
public:
    string convert(string s, int nRows) {
        if(nRows <= 1 || s == "") return s;
        int len = s.length();
		vector<int> index(2*nRows-2);
		int t = 0;
		for(int i = 0; i < nRows; ++i) index[t++] = i;
		for(int j = nRows-2; j > 0; --j) index[t++] = j;
		string s2[nRows];
        int k = 0, m = 2*(nRows-1);  
        while(k < len){
			s2[index[k % m]].push_back(s[k]);
			k++;
        }
        string s3;
        for(int i = 0; i < nRows; ++i) {
            s3 += s2[i];
        }
        return s3;
    }
};