Leetcode 1074 元素和为目标值的子矩阵数量 前缀和

题目链接

二维前缀和暴力思路，时间复杂度 $O(n^2\times m^2)$

class Solution {
public:
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int s[110][110];
        int n = matrix.size(), m = matrix[0].size();
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                s[i+1][j+1] = matrix[i][j] + s[i+1][j] + s[i][j+1] - s[i][j];
        int res = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                for(int x = 1; x <= i; x++)
                {
                    for(int y = 1; y <= j; y++)
                    {
                        int t = s[i][j] - s[x-1][j] - s[i][y-1] + s[x-1][y-1];
                        if(t == target) res++;
                    }
                }
            }
        }
        return res;
    }
};

哈希表+前缀和线性优化，最优的一个做法的时间复杂度$O(min(n,m)^2\times max(n,m))$

class Solution {
public:
    int solve(vector<int>& num, int target)
    {
        int cnt = 0, pre = 0, len = num.size();
        unordered_map<int, int> hash;
        hash[0] = 1;
        for(int i = 0; i < len; i++)
        {
            pre += num[i];
            if(hash.find(pre - target) != hash.end())
                cnt += hash[pre-target];
            hash[pre]++;
        }
        return cnt;
    }
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int n = matrix.size(), m = matrix[0].size();
        int x = min(n,m), y = max(n,m);
        int s[x+1][y+1], a[x+1][y+1];
        if(n > m)
        {
            for(int i = 0; i < n; i++)
                for(int j = 0; j < m; j++)
                    a[j][i] = matrix[i][j];
        }
        else{
            for(int i = 0; i < n; i++)
                for(int j = 0; j < m; j++)
                    a[i][j] = matrix[i][j];
        }
        int res = 0;
        for(int i = 0; i < x; i++)
        {
            vector<int> sum(y);
            for(int j = i; j < x; j++)
            {
                for(int k = 0; k < y; k++)
                    sum[k] += a[j][k]; 
                res += solve(sum, target);
            }
        }
        return res;
    }
};