题目描述:(中等难度)
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
题目解释:
String字符串在给定的行数下以锯齿状图形书写:(您可能希望以固定字体显示此模式以获得更好的可读性)
编写要接受字符串的代码,并使该转换给出若干行
题目解法:
1.我的解法。我的思路参考自HashMap,底层是数组,数组的长度由row决定,然后数组的每个子列都是链表,链表里按照要求存储数据,最后遍历数组 + 链表。先将String转成charArray,然后找规律分组,我这边分组是以第一行开始,到下一个第一行前结束,也就是每numRows + numRows - 2为一个循环,然后就是要注意如何找到数组的下标,在V字型向下的时候是正序,向上的时候是逆序。代码如下:
class Solution {
public String convert(String s, int numRows) {
if(numRows == 1) return s;
else {
char[] array = s.toCharArray();
List<List<Character>> list = new ArrayList<>();
for (int i = 0; i < numRows; i++){
List<Character> linkList = new LinkedList<>();
list.add(linkList);
}
int nums = numRows + numRows - 2;
List<Character> tempList = new LinkedList<>();
for(int i = 0; i < array.length; i++) {
if(i % nums == 0) {
tempList = list.get(0);
} else if(i % nums == numRows - 1) {
tempList = list.get(numRows - 1);
} else if(i % nums < numRows - 1) {
tempList = list.get(i % nums);
} else if(i % nums > numRows - 1) {
int x = (nums - i % nums);
tempList = list.get(x);
}
tempList.add(array[i]);
}
StringBuilder stringBuilder = new StringBuilder();
for(int i = 0; i < list.size(); i++) {
tempList = list.get(i);
for(int j = 0; j < tempList.size(); j++) {
stringBuilder.append(tempList.get(j));
}
}
return stringBuilder.toString();
}
}
}