leetcode-6-ZigZag Conversion
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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
思路:我觉得这就是一个找规律的题,写出一个字符串后,从第一行开始找,寻找每一行字母之间的关系
举例:
对于字符串s,设总行数为n=5
行数 起始字母下标 本行下一个字母的位置偏移数
one 0 (5-one)*2
上:0
two 1 下:(5-two)*2
上:(two-1)*2
three 2 下:(5-three)*2
上:(three-1)*2
four 3 下:(5-four)*2
上:(four-1)*2
five 4 下:(5-five)*2=0
上:(5-1)*2=8
代码如下:
1 class Solution { 2 public String convert(String s, int n) { 3 if(s.length()<=n||n == 1) return s; 4 StringBuilder sb=new StringBuilder(); 5 for(int row=1;row<=n;row++){ 6 int firstChar=row-1; 7 int down=(n-row)*2; 8 int up=(row-1)*2; 9 sb.append(s.charAt(firstChar)); 10 while(firstChar<s.length()){ 11 if(down+firstChar>=s.length())break; 12 else if(down!=0){ 13 sb.append(s.charAt(firstChar+down)); 14 firstChar=firstChar+down; 15 } 16 if(up+firstChar>=s.length()) break; 17 else if(up!=0) { 18 sb.append(s.charAt(firstChar+up)); 19 firstChar=firstChar+up; 20 } 21 } 22 } 23 return sb.toString(); 24 } 25 }
You are here!
Your runtime beats 99.22 % of java submissions.哈哈,感觉该没这么快吧。。。
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