5. ZigZag Conversion
难度: Medium
刷题内容
原题连接
- https://leetcode-cn.com/problems/zigzag-conversion
内容描述
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
解题方案
思路 1 **- 时间复杂度: O(len(s))*- 空间复杂度: O(len(s))***
需要将字符串s转换为按N排列，总共有numRows行，直接将字符串转换为N字形，然后输出, beats 74.18%
class Solution {
// 将字符串进行z子排列,行数为numRows
public String convert(String s, int numRows) {
// 思路：先转换为z字
List[] arr = new List[numRows]; // 保存每一行元素
for(int i = 0; i < numRows; i ++){
arr[i] = new ArrayList();
}
char[] chars = s.toCharArray();
for(int i = 0; i < chars.length;){
// 每次打印两列
for(int j = 0; j < numRows && i < chars.length; j++,i++){
List list = arr[j];
list.add(chars[i]);
}
for(int j = numRows - 1; j >= 0 && i < chars.length; j --){
if(j == numRows - 1 || j == 0){
arr[j].add(' ');
}else{
arr[j].add(chars[i]);
i++;
}
}
}
// 输出最终字符串
char[] result = new char[chars.length];
int index = 0;
for(int i = 0; i < numRows; i ++){
List list = arr[i];
for(int j = 0; j < list.size(); j ++){
if(' ' != (char)list.get(j)){
result[index++] = (char) list.get(j);
}
}
}
return new String(result);
}
}
