做法1

质数/素数，1不是质数。
思路：维护最小索引和最大索引，一次遍历后返回两者差值。

class Solution {
public:
    bool isPrime(int num){
        if (num == 2) return true;
        else if (num == 1) return false;
        for (int i = 2; i < num; i++){
            if (num % i==0) return false;
        }
        return true;
    }
    int maximumPrimeDifference(vector<int>& nums) {
        int min_idx = 300000, max_idx = -1;
        for (int i = 0; i < nums.size(); i++){
            if (isPrime(nums[i])){
                min_idx = min(min_idx, i);
                max_idx = max(max_idx, i);
            }
        }
        return (max_idx - min_idx);

    }
};

做法2

数据范围，nums[i]在100以内，可以直接枚举出100以内的质数，另外题目求的是索引的最大距离，因此找到第一个质数位置，后面找到新的质数位置直接用i-first即可。

class Solution {
public:
    int maximumPrimeDifference(vector<int>& nums) {
        unordered_set<int> primes = {
            2,3,5,7,11,
            13,17,19,23,29,
            31,37,41,43,47,
            53,59,61,67,71,
            73,79,83,89,97
        };
        int n = nums.size();
        int first = -1, ans = 0;
        for (int i = 0; i < n; ++i) {
            if (primes.count(nums[i])) {
                if (first != -1) {
                    ans = max(ans, i - first);
                }
                else {
                    first = i;
                }
            }
        }
        return ans;
    }
};

python

class Solution:
    def maximumPrimeDifference(self, nums: List[int]) -> int:
        primes = {
            2, 3, 5, 7, 11,
            13, 17, 19, 23, 29,
            31, 37, 41, 43, 47,
            53, 59, 61, 67, 71,
            73, 79, 83, 89, 97
        }

        first, ans = -1, 0
        for i, num in enumerate(nums):
            if num in primes:
                if first != -1:
                    ans = max(ans, i - first)
                else:
                    first = i
        return ans