feat: add solutions to lcof problem: No.47

剑指 Offer 47. 礼物的最大价值

Author: yanglbme

lcof/面试题47. 礼物的最大价值/README.md

@@ -40,6 +40,8 @@ $$
 
 时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为棋盘的行数和列数。
 
+我们注意到 $f[i][j]$ 只与 $f[i-1][j]$ 和 $f[i][j-1]$ 有关，因此我们可以仅用两行数组 $f[2][n+1]$ 来存储状态，从而将空间复杂度优化到 $O(n)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -55,6 +57,17 @@ class Solution:
         return f[m][n]
 ```
 
+```python
+class Solution:
+    def maxValue(self, grid: List[List[int]]) -> int:
+        m, n = len(grid), len(grid[0])
+        f = [[0] * (n + 1) for _ in range(2)]
+        for i, row in enumerate(grid, 1):
+            for j, v in enumerate(row, 1):
+                f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + v
+        return f[m & 1][n]
+```
+
 ### **Java**
 
 ```java
@@ -72,6 +85,21 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int maxValue(int[][] grid) {
+        int m = grid.length, n = grid[0].length;
+        int[][] f = new int[2][n + 1];
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                f[i & 1][j] = Math.max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
+            }
+        }
+        return f[m & 1][n];
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -90,6 +118,22 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    int maxValue(vector<vector<int>>& grid) {
+        int m = grid.size(), n = grid[0].size();
+        vector<vector<int>> f(2, vector<int>(n + 1, 0));
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
+            }
+        }
+        return f[m & 1][n];
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -115,6 +159,29 @@ func max(a, b int) int {
 }
 ```
 
+```go
+func maxValue(grid [][]int) int {
+	m, n := len(grid), len(grid[0])
+	f := make([][]int, 2)
+	for i := range f {
+		f[i] = make([]int, n+1)
+	}
+	for i := 1; i <= m; i++ {
+		for j := 1; j <= n; j++ {
+			f[i&1][j] = max(f[i&1^1][j], f[i&1][j-1]) + grid[i-1][j-1]
+		}
+	}
+	return f[m&1][n]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -125,17 +192,33 @@ func max(a, b int) int {
 var maxValue = function (grid) {
     const m = grid.length;
     const n = grid[0].length;
-    let dp = new Array(m + 1);
-    for (let i = 0; i < m + 1; ++i) {
-        dp[i] = new Array(n + 1).fill(0);
+    const f = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
+        }
     }
-    for (let i = 1; i < m + 1; ++i) {
-        for (let j = 1; j < n + 1; ++j) {
-            dp[i][j] =
-                Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1];
+    return f[m][n];
+};
+```
+
+```js
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maxValue = function (grid) {
+    const m = grid.length;
+    const n = grid[0].length;
+    const f = new Array(2).fill(0).map(() => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            f[i & 1][j] =
+                Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
+                grid[i - 1][j - 1];
         }
     }
-    return dp[m][n];
+    return f[m & 1][n];
 };
 ```
 
@@ -155,6 +238,22 @@ function maxValue(grid: number[][]): number {
 }
 ```
 
+```ts
+function maxValue(grid: number[][]): number {
+    const m = grid.length;
+    const n = grid[0].length;
+    const f = Array.from({ length: 2 }, _ => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            f[i & 1][j] =
+                Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
+                grid[i - 1][j - 1];
+        }
+    }
+    return f[m & 1][n];
+}
+```
+
 ### **Rust**
 
 ```rust

lcof/面试题47. 礼物的最大价值/Solution.cpp

@@ -2,12 +2,12 @@ class Solution {
 public:
     int maxValue(vector<vector<int>>& grid) {
         int m = grid.size(), n = grid[0].size();
-        vector<vector<int>> f(m + 1, vector<int>(n + 1, 0));
+        vector<vector<int>> f(2, vector<int>(n + 1, 0));
         for (int i = 1; i <= m; ++i) {
             for (int j = 1; j <= n; ++j) {
-                f[i][j] = max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
+                f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
             }
         }
-        return f[m][n];
+        return f[m & 1][n];
     }
 };

lcof/面试题47. 礼物的最大价值/Solution.go

@@ -1,15 +1,15 @@
 func maxValue(grid [][]int) int {
 	m, n := len(grid), len(grid[0])
-	f := make([][]int, m+1)
+	f := make([][]int, 2)
 	for i := range f {
 		f[i] = make([]int, n+1)
 	}
 	for i := 1; i <= m; i++ {
 		for j := 1; j <= n; j++ {
-			f[i][j] = max(f[i-1][j], f[i][j-1]) + grid[i-1][j-1]
+			f[i&1][j] = max(f[i&1^1][j], f[i&1][j-1]) + grid[i-1][j-1]
 		}
 	}
-	return f[m][n]
+	return f[m&1][n]
 }
 
 func max(a, b int) int {

lcof/面试题47. 礼物的最大价值/Solution.java

@@ -1,12 +1,12 @@
 class Solution {
     public int maxValue(int[][] grid) {
         int m = grid.length, n = grid[0].length;
-        int[][] f = new int[m + 1][n + 1];
+        int[][] f = new int[2][n + 1];
         for (int i = 1; i <= m; ++i) {
             for (int j = 1; j <= n; ++j) {
-                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
+                f[i & 1][j] = Math.max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
             }
         }
-        return f[m][n];
+        return f[m & 1][n];
     }
 }

lcof/面试题47. 礼物的最大价值/Solution.js

@@ -5,11 +5,13 @@
 var maxValue = function (grid) {
     const m = grid.length;
     const n = grid[0].length;
-    const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
+    const f = new Array(2).fill(0).map(() => new Array(n + 1).fill(0));
     for (let i = 1; i <= m; ++i) {
         for (let j = 1; j <= n; ++j) {
-            f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
+            f[i & 1][j] =
+                Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
+                grid[i - 1][j - 1];
         }
     }
-    return f[m][n];
+    return f[m & 1][n];
 };

lcof/面试题47. 礼物的最大价值/Solution.py

@@ -1,8 +1,8 @@
 class Solution:
     def maxValue(self, grid: List[List[int]]) -> int:
         m, n = len(grid), len(grid[0])
-        f = [[0] * (n + 1) for _ in range(m + 1)]
+        f = [[0] * (n + 1) for _ in range(2)]
         for i, row in enumerate(grid, 1):
             for j, v in enumerate(row, 1):
-                f[i][j] = max(f[i - 1][j], f[i][j - 1]) + v
-        return f[m][n]
+                f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + v
+        return f[m & 1][n]

lcof/面试题47. 礼物的最大价值/Solution.ts

@@ -1,11 +1,13 @@
 function maxValue(grid: number[][]): number {
     const m = grid.length;
     const n = grid[0].length;
-    const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
+    const f = Array.from({ length: 2 }, _ => new Array(n + 1).fill(0));
     for (let i = 1; i <= m; ++i) {
         for (let j = 1; j <= n; ++j) {
-            f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
+            f[i & 1][j] =
+                Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
+                grid[i - 1][j - 1];
         }
     }
-    return f[m][n];
+    return f[m & 1][n];
 }