【ZigZag Conversion】cpp

题目：

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".

代码：

class Solution {
public:
    string convert(string s, int numRows) {
            const int len = s.size();
            if ( len<numRows || numRows== ) return s;
            vector<char> ret;
            const int INTERVAL = *numRows-;
            for ( int i=; i<numRows; ++i )
            {
                int interval = *(numRows-i)-;
                for ( int j=i; j<len; interval=INTERVAL-interval)
                {
                    ret.push_back(s[j]);
                    j = (interval==INTERVAL||interval==) ? j+INTERVAL : j+interval;
                }
            }
            return string(ret.begin(),ret.end());
    }
};

tips：

找到ZigZag每行元素在元字符串s中间隔大小的规律。

=======================================

第二次过这道题，思路还是第一次的思路，但是代码不如第一次简洁了。

class Solution {
public:
    string convert(string s, int numRows) {
            vector<char> ret;
            if ( numRows== ) return s;
            const int interval = *(numRows-);
            for ( int i=; i<numRows; ++i )
            {
                int j = i;
                int preInterval = *i;
                while ( j<s.size() )
                {
                    ret.push_back(s[j]);
                    if ( preInterval!= && preInterval!=interval )
                    {
                        j = j + interval - preInterval;
                        preInterval = interval - preInterval;
                    }
                    else
                    {
                        j = j + interval;
                    }
                }
            }
            return string(ret.begin(),ret.end());
    }
};