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 s, int numRows);
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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
Example 3:
Input: s = "A", numRows = 1
Output: "A"
Constraints:
1 <= s.length <= 1000
s consists of English letters (lower-case and upper-case), ',' and '.'.
1 <= numRows <= 1000
C++ Solution
没有想到比较好的方法，这题的思路是先找到竖的集合以及斜的集合，然后通过竖的可以获取到最终的头和尾，竖和斜的集合进行处理就可以组成字符串的中间部分了。
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#include <iostream>
#include <string>
using namespace std;
class Solution {
public:
string convert(string s, int numRows) {
vector<string> zags;
vector<string> zigs;
vector<string> middle_vec;
string header, middle, end;
int col_index = 0;
int zig_index = 0;
if (numRows == 1) {
return s;
}
for (int i = 0; i < s.length(); i += 2 * (numRows - 1)) {
zags.push_back(s.substr(i, numRows));
}
for (int i = numRows; i < s.length(); i += 2 * (numRows - 1)) {
string tmp = s.substr(i, numRows - 2);
reverse(tmp.begin(), tmp.end());
zigs.push_back(tmp);
}
while (col_index < zags.size() || zig_index < zigs.size()) {
if (col_index < zags.size()) {
header.push_back(zags[col_index][0]);
if (zags[col_index].size() == numRows) {
end.push_back(zags[col_index][numRows - 1]);
}
if (zags[col_index].size() == numRows) {
middle_vec.push_back(zags[col_index].substr(1, numRows - 2));
} else {
middle_vec.push_back(zags[col_index].substr(1));
}
col_index++;
}
if (zig_index < zigs.size()) {
string tmp = zigs[zig_index];
middle_vec.push_back(zigs[zig_index++]);
}
}
for (int i = 0; i < numRows - 2; i++) {
int index = 0;
for (auto str : middle_vec) {
if (str.size() == numRows - 2) {
middle.push_back(str[i]);
} else {
if (index % 2 == 1){
if (i > numRows - 2 - str.size() - 1) {
middle.push_back(str[i - (numRows - 2 - str.size())]);
}
} else {
if (i < str.size()) {
middle.push_back(str[i]);
}
}
}
index++;
}
}
return header + middle + end;
}
};
