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：z字型填一个矩阵，时间复杂度为O(n^2)，注意边界值
1 class Solution { 2 public: 3 string convert(string s, int numRows) { 4 if(s.size()==0 || s.size()<=numRows || numRows<=1)return s;//注意边界值 5 int numCols=s.size()*2/(numRows*2-2)+1; 6 char **arr=(char**)malloc(sizeof(char*)*numRows); 7 for(int i=0; i<numRows; ++i)//初始化矩阵 8 { 9 arr[i]=(char*)malloc(sizeof(char)*numCols); 10 for(int j=0; j<numCols; ++j)arr[i][j]=‘\0‘; 11 } 12 int k=0; 13 for(int j=0; j<numCols; j=j+2)//填矩阵 14 { 15 for(int i=0; i<numRows && k<s.size(); ++i)arr[i][j]=s[k++]; 16 if(j+1<numCols) 17 { 18 for(int i=numRows-2; i>=1 && k<s.size(); --i)arr[i][j+1]=s[k++]; 19 } 20 } 21 string str; 22 for(int i=0; i<numRows; ++i)//遍历矩阵得到结果 23 { 24 for(int j=0; j<numCols; ++j) 25 { 26 if(arr[i][j]!=‘\0‘)str=str+arr[i][j]; 27 } 28 } 29 for(int i=0; i<numRows; ++i)delete []arr[i];//释放内存 30 delete []arr; 31 return str; 32 } 33 };
