题目(来自leetcode网站):
实现了python版本
实现了C++版本
题目含义为,将输入字符串 首先根据numRows大小进行 列的 Z 字形 排列 后,把每一行直接拼接起来输出;
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
解答:
#我的解法比较low 是根据人思考的逻辑来的,希望高手大佬指点一二
#python版本
class Solution:
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
a=[]
if len(s)<=numRows or numRows==1:
return s
for i in range(numRows):
a.append([])
i=0
j=0
while(i<len(s)):
for k in range(numRows):
if j == 0:
a[k].append(s[i])
i+=1
if i >=len(s):
break
else:
a[numRows-j-1].append(s[i])
i+=1
break
j+=1
if j % (numRows-1)==0:
j = 0
b=""
for i in range(numRows):
print(a[i])
for j in range(len(a[i])):
b+=a[i][j]
return b
#C++版本
class Solution {
public:
string convert(string s, int numRows) {
//string** a=new string*[numRows];
string a[numRows];
if(s.size()<=numRows || numRows ==1){
return s;
}
int i=0;
int j=0;
cout<<a[0]<<endl;
while(i < s.size()){
for(int k=0;k<numRows;k++){
if(j==0){
a[k]+=s[i];
i++;
if(i >=s.size()){
break;
}
}else{
a[numRows-j-1]+=s[i];
i++;
break;
}
}
j++;
if(j%(numRows-1)==0){
j=0;
}
}
string out;
for(int i=0;i<numRows;i++){
out+=a[i];
}
return out;
}
};