# 1074.Number-of-Submatrices-That-Sum-to-Target

## 1074. Number of Submatrices That Sum to Target

## 题目地址

<https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/>

## 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
```

## 代码

### Approach #1 Horizontal 1D Prefix Sum

Time: O(R^2 \* C) && Space: O(RC)

```java
class Solution {
  public int numSubmatrixSumTarget(int[][] matrix, int target) {
        int r = matrix.length;
    int c = matrix[0].length;

    int[][] ps = new int[r + 1][c + 1];
    for (int i = 1; i < r + 1; i++) {
      for (int j = 1; j < c + 1; j++) {
        ps[i][j] = ps[i - 1][j] + ps[i][j - 1] -ps[i - 1][j - 1] + matrix[i - 1][j - 1];
      }
    }

    int count = 0, currSum;
    Map<Integer, Integer> h = new HashMap();

    for (int r1 = 1; r1 < r + 1; r1++) {
      for (int r2 = r1; r2 < r + 1; r2++) {
        h.clear();
        h.put(0, 1);
        for (int col = 1; col < c + 1; col++) {
          currSum = ps[r2][col] - ps[r1 - 1][col];
          count += h.getOrDefault(currSum - target, 0);
          h.put(currSum, h.getOrDefault(currSum, 0) + 1);
        }
      }
    }

    return count;
  }
}
```


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