LeetCode 1071. Greatest Common Divisor of Strings (Java版; Easy)

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LeetCode 1071. Greatest Common Divisor of Strings (Java版; Easy)

题目描述

For strings S and T, we say "T divides S" if and only if S = T + ... + T  (T concatenated with itself 1 or more times)

Return the largest string X such that X divides str1 and X divides str2.

 

Example 1:

Input: str1 = "ABCABC", str2 = "ABC"
Output: "ABC"
Example 2:

Input: str1 = "ABABAB", str2 = "ABAB"
Output: "AB"
Example 3:

Input: str1 = "LEET", str2 = "CODE"
Output: ""
 

Note:

1 <= str1.length <= 1000
1 <= str2.length <= 1000
str1[i] and str2[i] are English uppercase letters.

第一次做; 暴力; 核心:1)如果两个字符串str1和str2的最大公因字符串是s, 设该字符串长度为n, 那么str1.substring(0,n)一定等于s, str2.substring(0,n)一定等于s

class Solution {
    public String gcdOfStrings(String str1, String str2) {
        //暴力
        int n1 = str1.length(), n2 = str2.length();
        int min = Math.min(n1, n2);
        for(int i=min; i>=1; i--){
            //当前长度既是n1的因数又是n2的因数时判断是否是gcd; tip:这里的i其实是n1和n2的因数, 但不一定是最大公因数
            if(n1%i==0 && n2%i==0){
                boolean flag = check(str1, str2.substring(0,i)) && check(str2, str1.substring(0,i));
                if(flag)
                    return str1.substring(0,i);
            }
        }
        return "";
    }

    private boolean check(String str, String s){
        int n = str.length()/s.length();
        StringBuilder sb = new StringBuilder();
        for(int i=0; i<n; i++){
            sb.append(s);
        }
        return sb.toString().equals(str);
    }
}

第一次做; 直接看的题解; 数学法, 数学才是王道啊!

class Solution {
    public String gcdOfStrings(String str1, String str2) {
        if(!(str1+str2).equals(str2+str1))
            return "";
        return str1.substring(0, gcd(str1.length(), str2.length()));
    }

    private int gcd(int a, int b){
        return b==0? a : gcd(b, a%b);
    }
}

最大公因数的求法, 辗转相除法, 核心: 1)辗转 2)取模

//递归版
private int gcd(int a, int b){
    return b==0? a: gcd(b, a%b);
}

//迭代版
private int gcd(int a, int b){
       while(b != 0){
           int tmp = b;
           b = a%b;
           a = tmp;
       }
       return a;
   }
   
/*
12 9 3
9 3 3
3 0 3

12 7 1
7  5 1
5  2 1
2  1 1
1  0 1
*/