【LeetCode】312. Burst Balloons-优快云博客

探讨了一道经典动态规划问题，目标是通过策略性地爆破带有数值的气球来获得最大金币数。文章提供了四种不同实现方案，包括递归、自底向上动态规划、记忆化搜索和优化的空间复杂度解决方案。

题目：

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

   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

题解：

Solution 1 (TLE)

class Solution {
public:
    void helper(vector<int> nums, int cur, int& res) {
        if(nums.size() == 2) {
            if(cur > res) res = cur;
            return;
        }
        for(int i=1; i<nums.size()-1; ++i) {
            int tmp = nums[i];
            cur += nums[i-1]*nums[i]*nums[i+1];
            nums.erase(nums.begin()+i);
            helper(nums, cur, res);
            nums.insert(nums.begin()+i,tmp);
            cur -= nums[i-1]*nums[i]*nums[i+1];
        }
    }
    int maxCoins(vector<int> nums) {
        nums.insert(nums.begin(),1);
        nums.push_back(1);
        int res = INT_MIN;
        helper(nums, 0, res);
        return res;    
    }
};

Solution 2 ()

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        nums.insert(nums.begin(), 1);
        nums.push_back(1);
        vector<vector<int> > dp(nums.size(), vector<int>(nums.size() , 0));
        for (int len = 1; len <= n; ++len) {
            for (int left = 1; left <= n - len + 1; ++left) {
                int right = left + len - 1;
                for (int k = left; k <= right; ++k) {
                    dp[left][right] = max(dp[left][right], nums[left - 1] * nums[k] * nums[right + 1] + dp[left][k - 1] + dp[k + 1][right]);
                }
            }
        }
        return dp[1][n];
  /*    for (int len = 2; len <= n+1; ++len) {
            for (int left = 0; left <= n - len + 1; ++left) {
                int right = left + len;
                for (int k = left+1; k < right; ++k) {
                    dp[left][right] = max(dp[left][right], nums[left] * nums[k] * nums[right] + dp[left][k] + dp[k][right]);
                }
            }
        }
        return dp[0][n+1]; */
    }
};

Solution 3 ()

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        nums.insert(nums.begin(), 1);
        nums.push_back(1);
        vector<vector<int> > dp(nums.size(), vector<int>(nums.size() , 0));
        return burst(nums, dp, 1 , n);
    }
    int burst(vector<int> &nums, vector<vector<int> > &dp, int left, int right) {
        if (left > right) return 0;
        if (dp[left][right] > 0) return dp[left][right];
        int res = 0;
        for (int k = left; k <= right; ++k) {
            res = max(res, nums[left - 1] * nums[k] * nums[right + 1] + burst(nums, dp, left, k - 1) + burst(nums, dp, k + 1, right));
        }
        dp[left][right] = res;
        return res;
    }
};

Solution 4 ()

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        nums.insert(nums.begin(),1);
        nums.push_back(1);
        const auto N=nums.size();
        vector<int> m(N*N);
        for(size_t l=2;l<N;l++)
        {
            for(size_t i=0;i+l<N;i++)
            {
                const size_t j=i+l;
                int v=0;
                for(size_t k=i+1;k<j;k++)
                {
                    v=max(v,nums[i]*nums[k]*nums[j]+m[i*N+k]+m[k*N+j]);
                }
                m[i*N+j]=v;
            }
        }
        return m[N-1];
    }
};