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feat: add solutions to lc problem: No.3363 (doocs#4625)
No.3363.Find the Maximum Number of Fruits Collected
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‎solution/3300-3399/3363.Find the Maximum Number of Fruits Collected/README.md

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<!-- solution:start -->
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### 方法一
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### 方法一：动态规划
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根据题目描述，从 $(0, 0)$ 出发的小朋友要想在 $n - 1$ 步后到达 $(n - 1, n - 1)$，那么他只能走主对角线上的房间 $(i, i)$，即 $i = 0, 1, \ldots, n - 1$。而从 $(0, n - 1)$ 出发的小朋友只能走主对角线以上的房间，而从 $(n - 1, 0)$ 出发的小朋友只能走主对角线以下的房间。这意味着三个小朋友除了在 $(n - 1, n - 1)$ 到达终点外，其他房间都不会有多个小朋友重复进入。
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我们可以用动态规划的方式，计算从 $(0, n - 1)$ 和 $(n - 1, 0)$ 出发的小朋友达到 $(i, j)$ 时，能收集到的水果数。定义 $f[i][j]$ 表示小朋友到达 $(i, j)$ 时能收集到的水果数。
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对于从 $(0, n - 1)$ 出发的小朋友，状态转移方程为：
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$$
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f[i][j] = \max(f[i - 1][j], f[i - 1][j - 1], f[i - 1][j + 1]) + \text{fruits}[i][j]
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$$
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注意，只有当 $j + 1 < n$ 时，$f[i - 1][j + 1]$ 才是有效的。
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对于从 $(n - 1, 0)$ 出发的小朋友，状态转移方程为：
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$$
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f[i][j] = \max(f[i][j - 1], f[i - 1][j - 1], f[i + 1][j - 1]) + \text{fruits}[i][j]
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$$
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同样，只有当 $i + 1 < n$ 时，$f[i + 1][j - 1]$ 才是有效的。
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最后，答案为 $\sum_{i=0}^{n-1} \text{fruits}[i][i] + f[n-2][n-1] + f[n-1][n-2]$，即主对角线上的水果数加上两个小朋友到达 $(n - 2, n - 1)$ 和 $(n - 1, n - 2)$ 时能收集到的水果数。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为房间的边长。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def maxCollectedFruits(self, fruits: List[List[int]]) -> int:
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n = len(fruits)
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f = [[-inf] * n for _ in range(n)]
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f[0][n - 1] = fruits[0][n - 1]
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for i in range(1, n):
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for j in range(i + 1, n):
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f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j]
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if j + 1 < n:
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f[i][j] = max(f[i][j], f[i - 1][j + 1] + fruits[i][j])
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f[n - 1][0] = fruits[n - 1][0]
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for j in range(1, n):
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for i in range(j + 1, n):
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f[i][j] = max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j]
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if i + 1 < n:
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f[i][j] = max(f[i][j], f[i + 1][j - 1] + fruits[i][j])
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return sum(fruits[i][i] for i in range(n)) + f[n - 2][n - 1] + f[n - 1][n - 2]
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```
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#### Java
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```java
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class Solution {
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public int maxCollectedFruits(int[][] fruits) {
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int n = fruits.length;
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final int inf = 1 << 29;
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int[][] f = new int[n][n];
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for (var row : f) {
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Arrays.fill(row, -inf);
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}
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f[0][n - 1] = fruits[0][n - 1];
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for (int i = 1; i < n; i++) {
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for (int j = i + 1; j < n; j++) {
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f[i][j] = Math.max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j];
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if (j + 1 < n) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j + 1] + fruits[i][j]);
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}
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}
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}
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f[n - 1][0] = fruits[n - 1][0];
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for (int j = 1; j < n; j++) {
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for (int i = j + 1; i < n; i++) {
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f[i][j] = Math.max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j];
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if (i + 1 < n) {
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f[i][j] = Math.max(f[i][j], f[i + 1][j - 1] + fruits[i][j]);
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}
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}
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}
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int ans = f[n - 2][n - 1] + f[n - 1][n - 2];
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for (int i = 0; i < n; i++) {
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ans += fruits[i][i];
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}
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int maxCollectedFruits(vector<vector<int>>& fruits) {
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int n = fruits.size();
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const int inf = 1 << 29;
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vector<vector<int>> f(n, vector<int>(n, -inf));
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f[0][n - 1] = fruits[0][n - 1];
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for (int i = 1; i < n; i++) {
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for (int j = i + 1; j < n; j++) {
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f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j];
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if (j + 1 < n) {
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f[i][j] = max(f[i][j], f[i - 1][j + 1] + fruits[i][j]);
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}
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}
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}
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f[n - 1][0] = fruits[n - 1][0];
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for (int j = 1; j < n; j++) {
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for (int i = j + 1; i < n; i++) {
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f[i][j] = max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j];
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if (i + 1 < n) {
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f[i][j] = max(f[i][j], f[i + 1][j - 1] + fruits[i][j]);
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}
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}
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}
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int ans = f[n - 2][n - 1] + f[n - 1][n - 2];
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for (int i = 0; i < n; i++) {
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ans += fruits[i][i];
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func maxCollectedFruits(fruits [][]int) int {
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n := len(fruits)
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const inf = 1 << 29
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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for j := range f[i] {
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f[i][j] = -inf
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}
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}
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f[0][n-1] = fruits[0][n-1]
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for i := 1; i < n; i++ {
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for j := i + 1; j < n; j++ {
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f[i][j] = max(f[i-1][j], f[i-1][j-1]) + fruits[i][j]
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if j+1 < n {
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f[i][j] = max(f[i][j], f[i-1][j+1]+fruits[i][j])
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}
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}
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}
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f[n-1][0] = fruits[n-1][0]
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for j := 1; j < n; j++ {
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for i := j + 1; i < n; i++ {
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f[i][j] = max(f[i][j-1], f[i-1][j-1]) + fruits[i][j]
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if i+1 < n {
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f[i][j] = max(f[i][j], f[i+1][j-1]+fruits[i][j])
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}
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}
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}
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ans := f[n-2][n-1] + f[n-1][n-2]
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for i := 0; i < n; i++ {
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ans += fruits[i][i]
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}
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return ans
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}
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```
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#### TypeScript
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```ts
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function maxCollectedFruits(fruits: number[][]): number {
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const n = fruits.length;
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const inf = 1 << 29;
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const f: number[][] = Array.from({ length: n }, () => Array(n).fill(-inf));
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f[0][n - 1] = fruits[0][n - 1];
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for (let i = 1; i < n; i++) {
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for (let j = i + 1; j < n; j++) {
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f[i][j] = Math.max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j];
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if (j + 1 < n) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j + 1] + fruits[i][j]);
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}
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}
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}
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f[n - 1][0] = fruits[n - 1][0];
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for (let j = 1; j < n; j++) {
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for (let i = j + 1; i < n; i++) {
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f[i][j] = Math.max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j];
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if (i + 1 < n) {
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f[i][j] = Math.max(f[i][j], f[i + 1][j - 1] + fruits[i][j]);
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}
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}
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}
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let ans = f[n - 2][n - 1] + f[n - 1][n - 2];
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for (let i = 0; i < n; i++) {
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ans += fruits[i][i];
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}
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return ans;
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn max_collected_fruits(fruits: Vec<Vec<i32>>) -> i32 {
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let n = fruits.len();
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let inf = 1 << 29;
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let mut f = vec![vec![-inf; n]; n];
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f[0][n - 1] = fruits[0][n - 1];
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for i in 1..n {
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for j in i + 1..n {
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f[i][j] = std::cmp::max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j];
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if j + 1 < n {
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f[i][j] = std::cmp::max(f[i][j], f[i - 1][j + 1] + fruits[i][j]);
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}
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}
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}
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f[n - 1][0] = fruits[n - 1][0];
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for j in 1..n {
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for i in j + 1..n {
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f[i][j] = std::cmp::max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j];
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if i + 1 < n {
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f[i][j] = std::cmp::max(f[i][j], f[i + 1][j - 1] + fruits[i][j]);
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}
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}
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}
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let mut ans = f[n - 2][n - 1] + f[n - 1][n - 2];
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for i in 0..n {
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ans += fruits[i][i];
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}
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ans
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}
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}
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```
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<!-- tabs:end -->

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