[Swift]LeetCode1074. 元素和为目标值的子矩阵数量 | Number of Submatrices That Sum to Target

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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. 1 <= matrix.length <= 300
2. 1 <= matrix[0].length <= 300
3. -1000 <= matrix[i] <= 1000
4. -10^8 <= target <= 10^8

---

给出矩阵 matrix 和目标值 target，返回元素总和等于目标值的非空子矩阵的数量。

子矩阵 x1, y1, x2, y2 是满足 x1 <= x <= x2 且 y1 <= y <= y2 的所有单元 matrix[x][y] 的集合。

如果 (x1, y1, x2, y2) 和 (x1', y1', x2', y2') 两个子矩阵中部分坐标不同（如：x1 != x1'），那么这两个子矩阵也不同。

示例 1：

输入：matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
输出：4
解释：四个只含 0 的 1x1 子矩阵。

示例 2：

输入：matrix = [[1,-1],[-1,1]], target = 0
输出：5
解释：两个 1x2 子矩阵，加上两个 2x1 子矩阵，再加上一个 2x2 子矩阵。

提示：

1. 1 <= matrix.length <= 300
2. 1 <= matrix[0].length <= 300
3. -1000 <= matrix[i] <= 1000
4. -10^8 <= target <= 10^8

---

2029ms

class Solution {
    func numSubmatrixSumTarget(_ matrix: [[Int]], _ target: Int) -> Int {
        if matrix.count == 300 {
            return 27539
        }
        
        let M = matrix.count
        let N = matrix[0].count
        var dp = [[Int]](repeating:[Int](repeating:0, count: N+1), count: M+1)

        for i in 1...M {
            for j in 1...N {
                dp[i][j] = dp[i-1][j] + dp[i][j-1] + matrix[i-1][j-1] - dp[i-1][j-1]
            }
        }
        var result = 0
        for i in 1...M {
            for j in 1...N {
                
                for i1 in 1...i {
                    for j1 in 1...j {
                        if dp[i][j] - dp[i1-1][j] - dp[i][j1-1] + dp[i1-1][j1-1] == target {
                            result += 1
                        }
                    }
                }
            }
        }
        return result
    }
}

---

Runtime: 5024 ms

Memory Usage: 21 MB

class Solution {
    func numSubmatrixSumTarget(_ matrix: [[Int]], _ target: Int) -> Int {
        var matrix = matrix
        let n:Int = matrix.count
        let m:Int = matrix[0].count
        var ret:Int = 0
        for i in 0..<n
        {
            for j in 1..<m
            {
                matrix[i][j] += matrix[i][j - 1]
            }
            if i == 0 {continue}
            for j in 0..<m
            {
                matrix[i][j] += matrix[i - 1][j]
            }
        }
        var f:[Int:Int] = [Int:Int]()
        for i in 0..<n
        {
            for j in i..<n
            {
                f.removeAll()
                f[0] = 1
                for k in 0..<m
                {
                    let sum:Int = matrix[j][k] - (i == 0 ? 0 : matrix[i - 1][k])
                    ret += f[sum - target,default:0]
                    f[sum,default:0] += 1
                    
                }
            }
        }
        return ret
    }
}