LeetCode #1074. Number of Submatrices That Sum to Target

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本文介绍了一种高效算法，用于计算给定矩阵中所有非空子矩阵的数目，这些子矩阵的元素之和等于指定的目标值。通过计算每行的前缀和，再枚举左右边界并使用哈希表，将二维问题转化为一维问题，从而显著提高了搜索效率。

题目描述：

Given a matrix, and a target, return the number of non-empty submatrices that sum to target.

A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2.

Two submatrices (x1, y1, x2, y2) and (x1', y1', x2', y2') are different if they have some coordinate that is different: for example, if x1 != x1'.

Example 1:

Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
Output: 4
Explanation: The four 1x1 submatrices that only contain 0.

Example 2:

Input: matrix = [[1,-1],[-1,1]], target = 0
Output: 5
Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.

Note:

1 <= matrix.length <= 300
1 <= matrix[0].length <= 300
-1000 <= matrix[i] <= 1000
-10^8 <= target <= 10^8

class Solution {
public:
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int m=matrix.size(), n=matrix[0].size();
        int result=0;
        for(int i=0;i<m;i++) // 计算每一行的前缀和
            for(int j=1;j<n;j++)
                matrix[i][j]+=matrix[i][j-1];
        // 枚举左右边界，由于i和j确定之后，利用哈希表变成一维的target sum
        for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                unordered_map<int,int> map; // 前缀和出现的次数
                map[0]=1;
                int prefix_sum=0;
                for(int k=0;k<m;k++)
                {   // 第k行中第i列到第j列的前缀和
                    if(i==0) prefix_sum+=matrix[k][j];
                    else prefix_sum+=matrix[k][j]-matrix[k][i-1];
                    if(map.count(prefix_sum-target))
                        result+=map[prefix_sum-target];
                    map[prefix_sum]++;
                }
            }
        }
        return result;
    }
};