LeetCode Counting Elements

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.

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.

Constraints:

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

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给定一个数组，如果数组中某个元素x和x+1都存在，则计数加一，问数组中共有多少个这样的x。

简单题，直接用set记录下有哪些数，再遍历看看x+1是否存在即可。代码如下：

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

本代码提交AC，用时8MS。