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zigzag-conversion2.cpp
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84 lines (61 loc) · 1.65 KB
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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 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
*/
#include <iostream>
#include <string>
#include <vector>
#include <stack>
class Solution {
public:
std::string convert(std::string s, int numRows) {
if(numRows == 1)
return s;
std::vector<std::stack<char>> vec(numRows);
int j=0; bool flag = 0;
for(int i=0; i<s.length(); i++)
{
vec[j].push(s[i]);
if(flag == 1) j--;
if(flag == 0) j++;
if(j == numRows){ flag = 1; j=numRows-2;}
if(j == -1){flag = 0; j=1;}
}
std::string result(s.length(),' ');
j = s.length();
for(int i= vec.size()-1; i>=0; i--)
{
while(!vec[i].empty())
{
result[j-1] = vec[i].top();
vec[i].pop();
j--;
}
}
return result;
}
};
int main()
{
Solution s;
std::string str = "PAYPALISHIRING";
std::string result = s.convert(str,4);
std::cout<<result<<'\n';
return 0;
}