标签：style blog io for div sp log on c
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;
}
};
标签：style blog io for div sp log on c
原文地址：http://www.cnblogs.com/liyangguang1988/p/3961253.html
