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".
這道題建立nRows個(gè)string存儲(chǔ)在vector中，然后便利string每個(gè)字符某筐，將字符加入到指向的vector中的string的末尾茸俭，指向位置先正向便利渴邦，再逆向遍歷菲嘴，循環(huán)直至傳入的string末尾爬虱。
string convert(string s, int numRows) {
if (numRows < 2) return s;
vector<string> v;
while (v.size() < numRows) {
string str;
v.push_back(str);
}
bool backward = true;
int i = 0;
int currentString = 0;
while (i < s.length()) {
v[currentString] += (s[i++]);
if (currentString == v.size() - 1) {
backward = false;
} else if (currentString == 0) {
backward = true;
}
currentString = backward ? (currentString + 1) : (currentString - 1);
}
string res;
for (int j = 0; j < v.size(); ++j) {
res.append(v[j]);
}
return res;
}
