0311. 稀疏矩阵的乘法

Author: progzc

go-leetcode/classify_algorithm/array/leetcode_0311_sparse-matrix-multiplication/leetcode_0311_sparse-matrix-multiplication.go

@@ -0,0 +1,90 @@
+package leetcode_0311_sparse_matrix_multiplication
+
+// 0311. 稀疏矩阵的乘法
+// https://leetcode.cn/problems/sparse-matrix-multiplication/
+
+// multiply 直接计算
+// 时间复杂度: O(m*n*k)
+// 空间复杂度: O(mn)
+func multiply(mat1 [][]int, mat2 [][]int) [][]int {
+	var (
+		m, k, n = len(mat1), len(mat1[0]), len(mat2[0])
+		ans     = make([][]int, m)
+	)
+	for i := 0; i < m; i++ {
+		ans[i] = make([]int, n)
+	}
+
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			for p := 0; p < k; p++ {
+				ans[i][j] += mat1[i][p] * mat2[p][j]
+			}
+		}
+	}
+	return ans
+}
+
+// multiply_2 直接计算（优化）
+// 时间复杂度: O(m*n*k)
+// 空间复杂度: O(mn)
+func multiply_2(mat1 [][]int, mat2 [][]int) [][]int {
+	var (
+		m, k, n = len(mat1), len(mat1[0]), len(mat2[0])
+		ans     = make([][]int, m)
+	)
+	for i := 0; i < m; i++ {
+		ans[i] = make([]int, n)
+	}
+
+	for i := 0; i < m; i++ {
+		for p := 0; p < k; p++ {
+			if mat1[i][p] == 0 {
+				continue
+			}
+			for j := 0; j < n; j++ {
+				ans[i][j] += mat1[i][p] * mat2[p][j]
+			}
+		}
+	}
+	return ans
+}
+
+// multiply_3 工程优化（空间换时间）
+// 时间复杂度: O(m*n)
+// 空间复杂度: O(mn)
+func multiply_3(mat1 [][]int, mat2 [][]int) [][]int {
+	var (
+		m, n = len(mat1), len(mat2[0])
+		ans  = make([][]int, m)
+	)
+	for i := 0; i < m; i++ {
+		ans[i] = make([]int, n)
+	}
+
+	noneZeroA := getNoneZeroMat(mat1)
+	noneZeroB := getNoneZeroMat(mat2)
+	for _, m1 := range noneZeroA {
+		for _, m2 := range noneZeroB {
+			// 这里这么判断的原因：mat1和mat2相乘，
+			// 只有mat1的数据的列数和mat2的行数相等才会进行计算
+			if m1[1] == m2[0] {
+				ans[m1[0]][m2[1]] += m1[2] * m2[2]
+			}
+		}
+	}
+	return ans
+}
+
+func getNoneZeroMat(matrix [][]int) [][]int {
+	m, n := len(matrix), len(matrix[0])
+	var ans [][]int
+	for i := 0; i < m; i++ {
+		for j := 0; j < n; j++ {
+			if matrix[i][j] != 0 {
+				ans = append(ans, []int{i, j, matrix[i][j]})
+			}
+		}
+	}
+	return ans
+}