检查数组是否是好的

You are given an integer array nums. We consider an array good if it is a permutation of an array base[n].

base[n] = [1, 2, ..., n - 1, n, n] (in other words, it is an array of length n + 1 which contains 1 to n - 1 exactly once, plus two occurrences of n). For example, base[1] = [1, 1] and base[3] = [1, 2, 3, 3].

Return true if the given array is good, otherwise return false.

Note: A permutation of integers represents an arrangement of these numbers.

 

Example 1:

Input: nums = [2, 1, 3]
Output: false
Explanation: Since the maximum element of the array is 3, the only candidate n for which this array could be a permutation of base[n], is n = 3. However, base[3] has four elements but array nums has three. Therefore, it can not be a permutation of base[3] = [1, 2, 3, 3]. So the answer is false.

Example 2:

Input: nums = [1, 3, 3, 2]
Output: true
Explanation: Since the maximum element of the array is 3, the only candidate n for which this array could be a permutation of base[n], is n = 3. It can be seen that nums is a permutation of base[3] = [1, 2, 3, 3] (by swapping the second and fourth elements in nums, we reach base[3]). Therefore, the answer is true.

Example 3:

Input: nums = [1, 1]
Output: true
Explanation: Since the maximum element of the array is 1, the only candidate n for which this array could be a permutation of base[n], is n = 1. It can be seen that nums is a permutation of base[1] = [1, 1]. Therefore, the answer is true.

Example 4:

Input: nums = [3, 4, 4, 1, 2, 1]
Output: false
Explanation: Since the maximum element of the array is 4, the only candidate n for which this array could be a permutation of base[n], is n = 4. However, base[4] has five elements but array nums has six. Therefore, it can not be a permutation of base[4] = [1, 2, 3, 4, 4]. So the answer is false.

 

Constraints:

• 1 <= nums.length <= 100
• 1 <= num[i] <= 200

/*
 * 思路：
 * 1. 使用 Java Stream 找到数组的最大值 max。
 * 2. 如果数组长度不等于 max + 1，则返回 false。
 * 3. 使用 Stream 生成 base 数组，包含 1 到 max，其中 max 重复两次。
 * 4. 排序 nums 和 base 数组，比较两者是否相等。
 */
import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public boolean isGoodArray(int[] nums) {
        int max = Arrays.stream(nums).max().orElse(0);
        if (nums.length != max + 1) {
            return false;
        }
        int[] base = IntStream.concat(IntStream.rangeClosed(1, max), IntStream.of(max)).toArray();
        Arrays.sort(nums);
        Arrays.sort(base);
        return Arrays.equals(nums, base);
    }
}

题解通过使用Counter类来统计nums中每个元素的出现次数。首先，通过比较Counter的长度和nums的长度减一来判断是否只有一个元素重复出现。其次，题解检查nums中的最大元素是否等于nums的长度减一，这是因为如果nums是一个好数组，那么它将包含所有从1到n的元素，并且n出现两次。最后，题解检验重复元素（即最大元素）出现了两次。如果这三个条件都满足，则返回true，否则返回false。

O(n)

O(n)