Leetcode:6. ZigZag Conversion
参考:http://blog.csdn.net/linhuanmars/article/details/21145039#reply,找规律
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 && j+size-2*i<s.length()) res.append(s.charAt(j+size-2*i)); } } return res.toString(); }
posted on 2017-12-19 10:13 Michael2397 阅读(98) 评论(0) 编辑 收藏 举报