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Given an integer array arr, count how many elements x there are, such that x + 1 is also in arr.

If there’re duplicates in arr, count them seperately.

Example 1:

Input: arr = [1,2,3]Output: 2Explanation: 1 and 2 are counted cause 2 and 3 are in arr.

Example 2:

Input: arr = [1,1,3,3,5,5,7,7]Output: 0Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.

Example 3:

Input: arr = [1,3,2,3,5,0]Output: 3Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.

Example 4:

Input: arr = [1,1,2,2]Output: 2Explanation: Two 1s are counted cause 2 is in arr.

Example 5:

Input: arr = [1,1,2]Output: 2Explanation: Both 1s are counted because 2 is in the array.

Constraints:

• 1 <= arr.length <= 1000
• 0 <= arr[i] <= 1000

题意：给出一个整数数组 arr， 对于元素 x ，只有当 x + 1 也在数组 arr 里时，才能记为 1 个数。如果数组 arr 里有重复的数，每个重复的数单独计算。

解法 哈希表

class Solution {
   public:    int countElements(vector
   
    & arr) {
           int cnt[1010] = {
   0}, ans = 0;        for (const int &v : arr) ++cnt[v];        for (int i = 0; i <= 1000; ++i)             if (cnt[i] && cnt[i + 1]) ans += cnt[i];        return ans;    }};
   

运行效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户内存消耗：7.6 MB, 在所有 C++ 提交中击败了48.15% 的用户

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