6. ZigZag Conversion
by Botao Xiao
Question:
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
Thinking:
class Solution {
public String convert(String s, int numRows) {
int len = 2 * numRows - 2;
StringBuilder[numRows] sbs = new StringBuilder[numRows];
for(int i = 0; i < numRows; i++){
sbs[i] = new StringBuilder();
}
int slen = s.length();
for(int i = 0; i < slen; i++){
int index = i % len;
if(index < len){
sbs[index].append(s.charAt(i).toString);
}else{
sbs[numRows - 1 - (index - numRows)].append(s.charAt(i).toString);
}
}
for(int i = 1; i < numRows; i++){
sbs[0].append(sbs[i].toString);
}
return sbs[0].toString;
}
}
二刷
- 创建numRows个StringBuilder数组。
- 用一个游标控制每次每个字符append在哪个位置。
- 通过判断由标的位置判断是向上还是向下。
class Solution {
public String convert(String s, int numRows) {
if(s == null || s.length() == 0 || numRows == 1) return s;
StringBuilder[] sbs = new StringBuilder[numRows];
for(int i = 0; i < numRows; i++)
sbs[i] = new StringBuilder();
int len = s.length();
int count = 0;
int index = 0;
boolean down = true;
while(index < len){
char c = s.charAt(index++);
sbs[count].append(c);
if(down) count++;
else count--;
if(count == numRows - 1 || count == 0) //此处要注意减1。
down = !down;
}
for(int i = 1; i < numRows; i++)
sbs[0].append(sbs[i].toString());
return sbs[0].toString();
}
}
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