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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思路：用一个集合存储元素，再遍历即可
class Solution:
def countElements(self, arr: List[int]) -> int:
myset = set(arr)
count = 0
for i in arr:
if i+1 in myset:
count += 1
return count
分析：
时间复杂度：O(N)
空间复杂度：O(N)