546. Remove Boxes--Hard

Given several boxes with different colors represented by different positive numbers.
You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), remove them and get k*k points.
Find the maximum points you can get.

Example 1:
Input:

[1, 3, 2, 2, 2, 3, 4, 3, 1]
Output:
23
Explanation:
[1, 3, 2, 2, 2, 3, 4, 3, 1]
----> [1, 3, 3, 4, 3, 1] (33=9 points)
----> [1, 3, 3, 3, 1] (11=1 points)
----> [1, 1] (33=9 points)
----> [] (22=4 points)
Note: The number of boxes n would not exceed 100.

1.思考

• 一开始想到的是DP，或者DFS搜索；
• 下面的实现代码使用的DFS搜索；
• 代码在本地运行时正确的，但是提交显示time limit exceeded，还要再优化；
  
  

2.实现

class Solution {
public:
    //This algorithm is right, but time limit exceeded.
    int mp = 0;

	int removeBoxes(vector<int>& boxes) {
		if (boxes.empty())
			return 0;
		DFS(boxes, 0);
		return mp;
	}

	void DFS(vector<int>& box, int point){
		if (box.empty())
			return;

		int len = box.size();
		int num = 0, color = box[0];
		vector<int> b = box;
		int i = 0, j = 0, p;

		while (i<len){
			b = box;
			num = 0;
			color = box[i];
			j = i;
			while (j<b.size()){
				if (b[j] == color){
					num++;
					b.erase(b.begin() + j);
					if (j < b.size())
						continue;
				}
				else
					break;
			}
			p = point + pow(num, 2);
			DFS(b, p);
			if (mp<p)
				mp = p;
			i += num;
		}

	}
    
};