Leetcode-2229

题目2229.Check if an Array Is Consecutive

难度：简单

Given an integer array nums, return true if nums is consecutive, otherwise return false.

An array is consecutive if it contains every number in the range [x, x + n - 1] (inclusive), where x is the minimum number in the array and n is the length of the array.

Example 1:

Input: nums = [1,3,4,2]
Output: true
Explanation:
The minimum value is 1 and the length of nums is 4.
All of the values in the range [x, x + n - 1] = [1, 1 + 4 - 1] = [1, 4] = (1, 2, 3, 4) occur in nums.
Therefore, nums is consecutive.

Example 2:

Input: nums = [1,3]
Output: false
Explanation:
The minimum value is 1 and the length of nums is 2.
The value 2 in the range [x, x + n - 1] = [1, 1 + 2 - 1], = [1, 2] = (1, 2) does not occur in nums.
Therefore, nums is not consecutive.

Example 3:

Input: nums = [3,5,4]
Output: true
Explanation:
The minimum value is 3 and the length of nums is 3.
All of the values in the range [x, x + n - 1] = [3, 3 + 3 - 1] = [3, 5] = (3, 4, 5) occur in nums.
Therefore, nums is consecutive.

Constraints:

• 1 <= nums.length <= 10⁵
• 0 <= nums[i] <= 10⁵

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/check-if-an-array-is-consecutive/
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解题思路

排序，新建a，判断是否在nums中
扩展思考: a所有元素（可能重复/不重复）是否在b中

解题代码

class Solution
{
public:
    bool isConsecutive(vector<int> &nums)
    {
        // if(nums == nullptr) return;
        int n = nums.size();
        // size
        sort(nums.begin(), nums.end());
        int min = nums[0];
        //最小值
        vector<int> vec;
        for (int i = min; i <= n + min - 1; i++)
        {
            vec.push_back(i);
        }
        int i = 0;
        while (i < vec.size())
        {
            if (vec[i] == nums[i])
                i++;
            else
                return false;
        }
        return true;
    }
};