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feat: add solutions to lcof problem: No.47
剑指 Offer 47. 礼物的最大价值
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‎lcof/面试题47. 礼物的最大价值/README.md

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@@ -40,6 +40,8 @@ $$
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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。
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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 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
@@ -55,6 +57,17 @@ class Solution:
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return f[m][n]
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```
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```python
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class Solution:
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def maxValue(self, grid: List[List[int]]) -> int:
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m, n = len(grid), len(grid[0])
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f = [[0] * (n + 1) for _ in range(2)]
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for i, row in enumerate(grid, 1):
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for j, v in enumerate(row, 1):
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f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + v
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return f[m & 1][n]
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```
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### **Java**
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```java
@@ -72,6 +85,21 @@ class Solution {
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}
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```
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```java
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class Solution {
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public int maxValue(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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int[][] f = new int[2][n + 1];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i & 1][j] = Math.max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
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}
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}
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return f[m & 1][n];
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}
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}
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```
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### **C++**
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```cpp
@@ -90,6 +118,22 @@ public:
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};
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```
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```cpp
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class Solution {
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public:
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int maxValue(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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vector<vector<int>> f(2, vector<int>(n + 1, 0));
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + grid[i - 1][j - 1];
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}
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}
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return f[m & 1][n];
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}
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};
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```
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### **Go**
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```go
@@ -115,6 +159,29 @@ func max(a, b int) int {
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}
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```
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```go
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func maxValue(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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f := make([][]int, 2)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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f[i&1][j] = max(f[i&1^1][j], f[i&1][j-1]) + grid[i-1][j-1]
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}
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}
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return f[m&1][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **JavaScript**
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```js
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var maxValue = function (grid) {
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const m = grid.length;
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const n = grid[0].length;
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let dp = new Array(m + 1);
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for (let i = 0; i < m + 1; ++i) {
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dp[i] = new Array(n + 1).fill(0);
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const f = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
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}
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}
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for (let i = 1; i < m + 1; ++i) {
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for (let j = 1; j < n + 1; ++j) {
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dp[i][j] =
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Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1];
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return f[m][n];
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};
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```
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```js
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/**
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* @param {number[][]} grid
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* @return {number}
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*/
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var maxValue = function (grid) {
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const m = grid.length;
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const n = grid[0].length;
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const f = new Array(2).fill(0).map(() => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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f[i & 1][j] =
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Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
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grid[i - 1][j - 1];
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}
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}
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return dp[m][n];
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return f[m & 1][n];
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};
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```
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}
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```
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```ts
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function maxValue(grid: number[][]): number {
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const m = grid.length;
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const n = grid[0].length;
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const f = Array.from({ length: 2 }, _ => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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f[i & 1][j] =
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Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
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grid[i - 1][j - 1];
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}
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}
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return f[m & 1][n];
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}
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```
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### **Rust**
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```rust

‎lcof/面试题47. 礼物的最大价值/Solution.cpp

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

‎lcof/面试题47. 礼物的最大价值/Solution.go

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@@ -1,15 +1,15 @@
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func maxValue(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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f := make([][]int, m+1)
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f := make([][]int, 2)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
9-
f[i][j] = max(f[i-1][j], f[i][j-1]) + grid[i-1][j-1]
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f[i&1][j] = max(f[i&1^1][j], f[i&1][j-1]) + grid[i-1][j-1]
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}
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}
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return f[m][n]
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return f[m&1][n]
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}
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func max(a, b int) int {

‎lcof/面试题47. 礼物的最大价值/Solution.java

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

‎lcof/面试题47. 礼物的最大价值/Solution.js

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var maxValue = function (grid) {
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const m = grid.length;
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const n = grid[0].length;
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const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
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const f = new Array(2).fill(0).map(() => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
1010
for (let j = 1; j <= n; ++j) {
11-
f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
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f[i & 1][j] =
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Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
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grid[i - 1][j - 1];
1214
}
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}
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return f[m][n];
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return f[m & 1][n];
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};

‎lcof/面试题47. 礼物的最大价值/Solution.py

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class Solution:
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def maxValue(self, grid: List[List[int]]) -> int:
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m, n = len(grid), len(grid[0])
4-
f = [[0] * (n + 1) for _ in range(m + 1)]
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f = [[0] * (n + 1) for _ in range(2)]
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for i, row in enumerate(grid, 1):
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for j, v in enumerate(row, 1):
7-
f[i][j] = max(f[i - 1][j], f[i][j - 1]) + v
8-
return f[m][n]
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f[i & 1][j] = max(f[i & 1 ^ 1][j], f[i & 1][j - 1]) + v
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return f[m & 1][n]

‎lcof/面试题47. 礼物的最大价值/Solution.ts

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function maxValue(grid: number[][]): number {
22
const m = grid.length;
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const n = grid[0].length;
4-
const f = Array.from({ length: m + 1 }, _ => new Array(n + 1).fill(0));
4+
const f = Array.from({ length: 2 }, _ => new Array(n + 1).fill(0));
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
7-
f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
7+
f[i & 1][j] =
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Math.max(f[(i & 1) ^ 1][j], f[i & 1][j - 1]) +
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grid[i - 1][j - 1];
810
}
911
}
10-
return f[m][n];
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return f[m & 1][n];
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}

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