一、題目描述
二、題解
方法一:
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s1 = sc.next();
String s2 = sc.next();
if (s1.length() != s2.length()) {
System.out.println(1);
return;
}
if (s1.equals(s2)) {
System.out.println(2);
return;
}
char[] S1 = s1.toCharArray();
char[] S2 = s2.toCharArray();
int cnt = 0;
for (int i = 0; i < s1.length(); i++) {
if (S1[i] == S2[i] || S1[i] + 32 == S2[i] || S1[i] - 32 == S2[i]) {
cnt++;
}
}
if (cnt == S1.length) {
System.out.println(3);
return;
}
else System.out.println(4);
}
}
複雜度分析
- 時間複雜度:O(n),
- 空間複雜度:O(1),
方法二:庫函數
s1.toLowerCase().equals(s2.toLowerCase()) && !s1.equals(s2)