Leetcode每日一题 —— 3718. 缺失的最小倍数

早上好佬友,今天来早了

class Solution:

    def missingMultiple(self, nums: List[int], k: int) -> int:

        nums = set(nums)

        x = k

        while k in nums:

            k += x

        return k

        

规模不大，标记出现的数字然后枚举即可。

class Solution {

public:

    int missingMultiple(vector<int>& nums, int k) {

        bool nMap[101];

        memset(nMap, 0, sizeof(nMap));

        for (int num : nums) {

            // 标记已经出现的数字

            nMap[num] = true;

        }

        int f = 1;

        while (f * k <= 100 && nMap[f * k]) {

            f++;

        }

        return f * k;

    }

};

算术评级 2 第 472 场周赛 Q1 难度分 1228

简简单单

class Solution:

    def missingMultiple(self, nums: List[int], k: int) -> int:

        nums = set(nums)

        for i in range(1,100//k+2):

            if i*k not in nums:

                return i*k

class Solution:

    def missingMultiple(self, nums: List[int], k: int) -> int:

        n = max(nums) // k

        for i in range(1, n + 1):

            if k * i not in nums:

                return k * i

        return k * (n + 1)