6. 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);
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
class Solution { public String convert(String s, int nRows) { if(s == null || s.length()==0 || nRows <=0) return ""; if(nRows == 1) return s; StringBuilder res = new StringBuilder(); int size = 2*nRows-2; for(int i=0;i<nRows;i++){ for(int j=i;j<s.length();j+=size){ res.append(s.charAt(j)); if(i != 0 && i != nRows - 1){//except the first row and the last row int temp = j+size-2*i; if(temp<s.length()) res.append(s.charAt(temp)); } } } return res.toString(); } }
2. simulation,numRows个stringbuffer,最后拼起来
class Solution { public String convert(String s, int numRows) { StringBuilder[] sb = new StringBuilder[numRows]; for (int i = 0; i < sb.length; i++) sb[i] = new StringBuilder(); int le = s.length(), ind = 0; char[] ch = s.toCharArray(); while(ind < le){ for(int i = 0; i < numRows && ind < le; i++){ sb[i].append(ch[ind++]);//纵向一列 } for(int i = numRows - 2; i > 0 && ind < le; i--){ sb[i].append(ch[ind++]);//处理非首尾行的情况 } } StringBuilder st = new StringBuilder(); for(int i = 0; i < numRows; i++) st.append(sb[i]); return st.toString(); } }