【leetcode】1442. Count Triplets That Can Form Two Arrays of Equal XOR

题目如下：

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

• a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
• b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Example 3:

Input: arr = [2,3]
Output: 0

Example 4:

Input: arr = [1,3,5,7,9]
Output: 3

Example 5:

Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8

Constraints:

• 1 <= arr.length <= 300
• 1 <= arr[i] <= 10^8

解题思路：本题的关键是找出j，j的取值范围是1~len(arr)-1。对于任意的j，首先计算出左边XOR值出现的次数，然后再依次计算右边的XOR值，再去左边的XOR里面找出相同的值出现了几次即可。

代码如下：

class Solution(object):
    def countTriplets(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        res = 0
        for i in range(1,len(arr)):
            dic_left = {}
            value = arr[i-1]
            dic_left[value] = 1
            for j in range(i-2,-1,-1):
                value = value ^ arr[j]
                dic_left[value] = dic_left.setdefault(value,0) + 1
            value = None
            for j in range(i,len(arr)):
                if value == None:value = arr[j]
                else:value = value ^ arr[j]
                if value in dic_left:
                    res += dic_left[value]
        return res