leetcode-6. ZigZag Conversion
题目:
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 text, int nRows); convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.
我看到这题的时候想的比较简单,就是用几个StringBuilder存一下每一行的值,然后组合在一起输出就行。这样做肯定要慢一些的。因为String方面的操作本身就很慢。如果可以用一个StringBuilder直接做肯定做肯定要快一些,或者用一个char[]做应该会更快。不过都是同一个数量级的没必要这么麻烦了。
我的解法
public class Solution {
public String
convert(String s,
int numRows) {
if(numRows <
2)
return s;
StringBuilder[] sbl =
new StringBuilder[numRows];
for(
int i =
0 ; i < sbl.length ; i++)
sbl[i] =
new StringBuilder();
boolean flag =
true;
for(
int i =
0,j =
0 ; i < s.length() ;){
if(flag)
sbl[j++].append(s.charAt(i++));
else
sbl[j--].append(s.charAt(i++));
if(j == numRows-
1 || j ==
0) flag = !flag;
}
String ret =
"";
for(StringBuilder sb : sbl)
ret += sb;
return ret;
}
}
用一个StringBuilder,然后指针计数的方法。
public String convert(String s,
int numRows) {
if(numRows <
2 || numRows >= s.
length())
return s;
StringBuilder sb = new StringBuilder(s.
length());
int origStep = numRows *
2 -
2;
int step = origStep;
for(
int i =
0; i < numRows; i++){
step = i == numRows -
1 ? origStep : origStep - i *
2;
int curr = i;
while(curr < s.
length()){
sb.append(s.charAt(curr));
curr +=
step;
int temp = Math.
abs(
step - origStep);
step = temp ==
0 ? origStep : temp;
}
}
return sb.toString();
}
解法来源
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