【LeetCode】 Counting Elements

Description

Given an integer array arr, count element x such that x + 1 is also in arr.

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

Constraints:

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

給予一正整數陣列arr，計算存在多少元素x當x + 1也在arr當中。

如果有重複的數在arr當中，將它們分開計算。

限制：

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

Example:

Example 1:

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


Example 4:

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

Solution

• 先跑一次迴圈，使用set紀錄哪些數字存在於陣列中。
• 再跑一次迴圈，根據剛剛的set計算count。

Code

class Solution {
public:
    int countElements(vector<int>& arr) {
        set<int> hash;
        for(int i = 0; i < arr.size(); i++)
            hash.insert(arr[i] - 1);
        int count = 0;
        for(int i = 0; i < arr.size(); i++)
            if(hash.find(arr[i]) != hash.end())
                count++;
        return count;
    }
};

tags: LeetCode C++