carloscn
commented
1 year ago Analysis
求两个数的最大公约数，可以使用更相减损法
思路：
1、如果a > b a = a - b; 2、如果b > a b = b - a; 3、假如a = b,则 a或 b是最大公约数； 4、如果a != b;则继续从一开始执行； 5、也就是说循环的判断条件为a != b,直到a = b时，循环结束。
举例说明：
a = 28 b = 21
a>b
则 a = a - b = 28 - 21 = 7
b = 21
b>a
则 b = b-a = 21 - 7 = 14
a = 7
b>a
则 b = b - a = 14 - 7 = 7
a = 7
此时 a = b =7
循环结束pub fn find_gcd(nums: Vec<i32>) -> i32
{
if nums.len() < 1 {
return 0;
}
let mut max_val:i32 = i32::MIN;
let mut min_val:i32 = i32::MAX;
for i in 0..nums.len() {
max_val = max_val.max(nums[i]);
min_val = min_val.min(nums[i]);
}
while max_val != min_val {
if max_val > min_val {
max_val -= min_val;
}
if max_val < min_val {
min_val -= max_val;
}
}
return max_val;
}
Description
Given an integer array nums, return the greatest common divisor of the smallest number and largest number in nums.
The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.
Example 1:
Input: nums = [2,5,6,9,10] Output: 2 Explanation: The smallest number in nums is 2. The largest number in nums is 10. The greatest common divisor of 2 and 10 is 2.
Example 2:
Input: nums = [7,5,6,8,3] Output: 1 Explanation: The smallest number in nums is 3. The largest number in nums is 8. The greatest common divisor of 3 and 8 is 1.
Example 3:
Input: nums = [3,3] Output: 3 Explanation: The smallest number in nums is 3. The largest number in nums is 3. The greatest common divisor of 3 and 3 is 3.
Constraints:
2 <= nums.length <= 1000 1 <= nums[i] <= 1000