feat(solution): add solution 0312 [Java]

Burst Balloons

Author: yanglbme

README.md

@@ -114,6 +114,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 0084 | [Largest Rectangle in Histogram](https://github.com/doocs/leetcode/tree/master/solution/0084.Largest%20Rectangle%20in%20Histogram) | `Array`, `Stack` |
 | 0145 | [Binary Tree Postorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/0145.Binary%20Tree%20Postorder%20Traversal) | `Stack`, `Tree` |
 | 0295 | [Find Median from Data Stream](https://github.com/doocs/leetcode/tree/master/solution/0295.Find%20Median%20from%20Data%20Stream) | `Heap`, `Design` |
+| 0312 | [Burst Balloons](https://github.com/doocs/leetcode/tree/master/solution/0312.Burst%20Balloons) | `Divide and Conquer`, `Dynamic Programming` |
 
 
 ## Contributions

solution/0312.Burst Balloons/README.md

@@ -0,0 +1,73 @@
+## 戳气球
+### 题目描述
+
+有 `n` 个气球，编号为 `0` 到 `n-1`，每个气球上都标有一个数字，这些数字存在数组 `nums` 中。
+
+现在要求你戳破所有的气球。每当你戳破一个气球 `i` 时，你可以获得 `nums[left] * nums[i] * nums[right]` 个硬币。 这里的 `left` 和 `right` 代表和 `i` 相邻的两个气球的序号。注意当你戳破了气球 `i` 后，气球 `left` 和气球 `right` 就变成了相邻的气球。
+
+求所能获得硬币的最大数量。
+
+**说明:**
+
+- 你可以假设 `nums[-1] = nums[n] = 1`，但注意它们不是真实存在的所以并不能被戳破。
+- 0 ≤ `n` ≤ 500, 0 ≤ `nums[i]` ≤ 100
+
+**示例:**
+```
+输入: [3,1,5,8]
+输出: 167 
+解释: nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
+     coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167
+
+```     
+
+### 解法
+数组 `f` 表示某范围内(开区间)戳破所有气球能获得的最大硬币。那么题意转换为求`f[0][n+1]`。
+
+假设最后一个戳破的气球为 `i`，那么：
+```
+f[0][n+1] = f[0][i] + f[i][n+1] + nums[i] * nums[0][n+1]。
+```
+
+利用记忆化搜索，遍历当最后一个戳破的气球为 `i` 时，求 `f[0][n+1]` 的最大值。
+
+```java
+class Solution {
+    
+    public int maxCoins(int[] nums) {
+        if (nums == null || nums.length == 0) {
+            return 0;
+        }
+        int n = nums.length;
+        int[][] f = new int[n + 2][n + 2];
+        for (int i= 0; i < n + 2; ++i) {
+            for (int j = 0; j < n + 2; ++j) {
+                f[i][j] = -1;
+            }
+        }
+        int[] bak = new int[n + 2];
+        bak[0] = bak[n + 1] = 1;
+        for (int i = 1; i < n + 1; ++i) {
+            bak[i] = nums[i - 1];
+        }
+        return dp(bak, f, 0, n + 1);
+    }
+    
+    private int dp(int[] nums, int[][] f, int x, int y) {
+        if (f[x][y] != -1) {
+            return f[x][y];
+        }
+        
+        f[x][y] = 0;
+        
+        //枚举最后一个戳破的气球的位置
+        for (int i = x + 1; i < y; ++i) {
+            f[x][y] = Math.max(f[x][y], nums[i] * nums[x] * nums[y] + dp(nums,f,  x, i) + dp(nums, f, i, y));
+        }
+        return f[x][y];
+        
+    }
+    
+    
+}
+```

solution/0312.Burst Balloons/Solution.java

@@ -0,0 +1,38 @@
+class Solution {
+    
+    public int maxCoins(int[] nums) {
+        if (nums == null || nums.length == 0) {
+            return 0;
+        }
+        int n = nums.length;
+        int[][] f = new int[n + 2][n + 2];
+        for (int i= 0; i < n + 2; ++i) {
+            for (int j = 0; j < n + 2; ++j) {
+                f[i][j] = -1;
+            }
+        }
+        int[] bak = new int[n + 2];
+        bak[0] = bak[n + 1] = 1;
+        for (int i = 1; i < n + 1; ++i) {
+            bak[i] = nums[i - 1];
+        }
+        return dp(bak, f, 0, n + 1);
+    }
+    
+    private int dp(int[] nums, int[][] f, int x, int y) {
+        if (f[x][y] != -1) {
+            return f[x][y];
+        }
+        
+        f[x][y] = 0;
+        
+        //枚举最后一个戳破的气球的位置
+        for (int i = x + 1; i < y; ++i) {
+            f[x][y] = Math.max(f[x][y], nums[i] * nums[x] * nums[y] + dp(nums,f,  x, i) + dp(nums, f, i, y));
+        }
+        return f[x][y];
+        
+    }
+    
+    
+}