和为目标值的子矩阵个数

class Solution:
    def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
        #求list的子list个数，子list的和=k，利用前缀和求解
        def sum(nums,k):
            count=0
            pre=0
            mp=Counter([0])
            for i in range(len(nums)):
                pre+=nums[i]
                if pre-k in mp:
                    count+=mp[pre-k]
                mp[pre]+=1
            return count
        re=0
        for m in range(len(matrix)):
            total=[0]*len(matrix[0])
            for n in range(m,len(matrix)):
                for c in range(len(matrix[0])):
                    total[c]+=matrix[n][c]
                re+=sum(total,target)
        return re
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/power-of-four/solution/4de-mi-by-leetcode-solution-b3ya/
来源：力扣（LeetCode）