modify code

algorithmzuo · algorithmzuo · commit cb86646344fc · 2022-04-06T15:00:00.000+08:00
diff --git a/src/class26/Code02_FibonacciProblem.java b/src/class26/Code02_FibonacciProblem.java
@@ -57,24 +57,27 @@ public static int[][] matrixPower(int[][] m, int p) {
 		int[][] t = m;// 矩阵1次方
 		for (; p != 0; p >>= 1) {
 			if ((p & 1) != 0) {
-				res = muliMatrix(res, t);
+				res = product(res, t);
 			}
-			t = muliMatrix(t, t);
+			t = product(t, t);
 		}
 		return res;
 	}
 
 	// 两个矩阵乘完之后的结果返回
-	public static int[][] muliMatrix(int[][] m1, int[][] m2) {
-		int[][] res = new int[m1.length][m2[0].length];
-		for (int i = 0; i < m1.length; i++) {
-			for (int j = 0; j < m2[0].length; j++) {
-				for (int k = 0; k < m2.length; k++) {
-					res[i][j] += m1[i][k] * m2[k][j];
+	public static int[][] product(int[][] a, int[][] b) {
+		int n = a.length;
+		int m = b[0].length;
+		int k = a[0].length; // a的列数同时也是b的行数
+		int[][] ans = new int[n][m];
+		for(int i = 0 ; i < n; i++) {
+			for(int j = 0 ; j < m;j++) {
+				for(int c = 0; c < k; c++) {
+					ans[i][j] += a[i][c] * b[c][j];
 				}
 			}
 		}
-		return res;
+		return ans;
 	}
 
 	public static int s1(int n) {
diff --git a/src/class26/Code03_ZeroLeftOneStringNumber.java b/src/class26/Code03_ZeroLeftOneStringNumber.java
@@ -78,23 +78,27 @@ public static int[][] matrixPower(int[][] m, int p) {
 		int[][] tmp = m;
 		for (; p != 0; p >>= 1) {
 			if ((p & 1) != 0) {
-				res = muliMatrix(res, tmp);
+				res = product(res, tmp);
 			}
-			tmp = muliMatrix(tmp, tmp);
+			tmp = product(tmp, tmp);
 		}
 		return res;
 	}
 
-	public static int[][] muliMatrix(int[][] m1, int[][] m2) {
-		int[][] res = new int[m1.length][m2[0].length];
-		for (int i = 0; i < m1.length; i++) {
-			for (int j = 0; j < m2[0].length; j++) {
-				for (int k = 0; k < m2.length; k++) {
-					res[i][j] += m1[i][k] * m2[k][j];
+	// 两个矩阵乘完之后的结果返回
+	public static int[][] product(int[][] a, int[][] b) {
+		int n = a.length;
+		int m = b[0].length;
+		int k = a[0].length; // a的列数同时也是b的行数
+		int[][] ans = new int[n][m];
+		for(int i = 0 ; i < n; i++) {
+			for(int j = 0 ; j < m;j++) {
+				for(int c = 0; c < k; c++) {
+					ans[i][j] += a[i][c] * b[c][j];
 				}
 			}
 		}
-		return res;
+		return ans;
 	}
 
 	public static void main(String[] args) {

Original file line number	Diff line number	Diff line change
`@@ -57,24 +57,27 @@ public static int[][] matrixPower(int[][] m, int p) {`
`57`	`57`	`int[][] t = m;// 矩阵1次方`
`58`	`58`	`for (; p != 0; p >>= 1) {`
`59`	`59`	`if ((p & 1) != 0) {`
`60`		`- res = muliMatrix(res, t);`
	`60`	`+ res = product(res, t);`
`61`	`61`	`}`
`62`		`- t = muliMatrix(t, t);`
	`62`	`+ t = product(t, t);`
`63`	`63`	`}`
`64`	`64`	`return res;`
`65`	`65`	`}`
`66`	`66`
`67`	`67`	`// 两个矩阵乘完之后的结果返回`
`68`		`- public static int[][] muliMatrix(int[][] m1, int[][] m2) {`
`69`		`- int[][] res = new int[m1.length][m2[0].length];`
`70`		`- for (int i = 0; i < m1.length; i++) {`
`71`		`- for (int j = 0; j < m2[0].length; j++) {`
`72`		`- for (int k = 0; k < m2.length; k++) {`
`73`		`- res[i][j] += m1[i][k] * m2[k][j];`
	`68`	`+ public static int[][] product(int[][] a, int[][] b) {`
	`69`	`+ int n = a.length;`
	`70`	`+ int m = b[0].length;`
	`71`	`+ int k = a[0].length; // a的列数同时也是b的行数`
	`72`	`+ int[][] ans = new int[n][m];`
	`73`	`+ for(int i = 0 ; i < n; i++) {`
	`74`	`+ for(int j = 0 ; j < m;j++) {`
	`75`	`+ for(int c = 0; c < k; c++) {`
	`76`	`+ ans[i][j] += a[i][c] * b[c][j];`
`74`	`77`	`}`
`75`	`78`	`}`
`76`	`79`	`}`
`77`		`- return res;`
	`80`	`+ return ans;`
`78`	`81`	`}`
`79`	`82`
`80`	`83`	`public static int s1(int n) {`
Original file line number	Diff line number	Diff line change
`@@ -78,23 +78,27 @@ public static int[][] matrixPower(int[][] m, int p) {`
`78`	`78`	`int[][] tmp = m;`
`79`	`79`	`for (; p != 0; p >>= 1) {`
`80`	`80`	`if ((p & 1) != 0) {`
`81`		`- res = muliMatrix(res, tmp);`
	`81`	`+ res = product(res, tmp);`
`82`	`82`	`}`
`83`		`- tmp = muliMatrix(tmp, tmp);`
	`83`	`+ tmp = product(tmp, tmp);`
`84`	`84`	`}`
`85`	`85`	`return res;`
`86`	`86`	`}`
`87`	`87`
`88`		`- public static int[][] muliMatrix(int[][] m1, int[][] m2) {`
`89`		`- int[][] res = new int[m1.length][m2[0].length];`
`90`		`- for (int i = 0; i < m1.length; i++) {`
`91`		`- for (int j = 0; j < m2[0].length; j++) {`
`92`		`- for (int k = 0; k < m2.length; k++) {`
`93`		`- res[i][j] += m1[i][k] * m2[k][j];`
	`88`	`+ // 两个矩阵乘完之后的结果返回`
	`89`	`+ public static int[][] product(int[][] a, int[][] b) {`
	`90`	`+ int n = a.length;`
	`91`	`+ int m = b[0].length;`
	`92`	`+ int k = a[0].length; // a的列数同时也是b的行数`
	`93`	`+ int[][] ans = new int[n][m];`
	`94`	`+ for(int i = 0 ; i < n; i++) {`
	`95`	`+ for(int j = 0 ; j < m;j++) {`
	`96`	`+ for(int c = 0; c < k; c++) {`
	`97`	`+ ans[i][j] += a[i][c] * b[c][j];`
`94`	`98`	`}`
`95`	`99`	`}`
`96`	`100`	`}`
`97`		`- return res;`
	`101`	`+ return ans;`
`98`	`102`	`}`
`99`	`103`
`100`	`104`	`public static void main(String[] args) {`