leetcode 312. Burst Balloons

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 iyou 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:

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

Example:

Input: [3,1,5,8]
Output: 167 
Explanation: 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

方法一：记忆化搜索

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int len = nums.size();
        vector<int> num(len + 2, 1);
        for (int i = 0; i < len; i++)
            num[i + 1] = nums[i];
        vector<vector<int> > memo(len + 2, vector<int>(len + 2, -1));
        return maxCoins(num, memo, 0, len + 1); 
    }
private:
    //表示（start， end）内的气球都点燃可以获得的最大钱数。
    int maxCoins(vector<int> &nums, vector<vector<int> > &memo, int start, int end) {
        if (memo[start][end] != -1)
            return memo[start][end];
        if (end - start == 1)
            memo[start][end] = 0;
        else {
            int income = 0;
            //分治。（start, i) 和 （i， end) 分别获得的最大钱数 + 点燃i的钱数
            for (int i = start + 1; i < end; i++) {
                income = max(income, maxCoins(nums, memo, start, i) + maxCoins(nums, memo, i, end) + nums[start] * nums[i] * nums[end]);
            }
            memo[start][end] = income;
        }
        return memo[start][end];
    }
};

方法二：动态规划(bottom up), 先解决只有一个气球，两个气球，...，

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int len = nums.size();
        vector<int> num(len + 2, 1);
        for (int i = 0; i < len; i++) {
            num[i + 1] = nums[i];
        }
        vector<vector<int> > dp(len + 2, vector<int>(len + 2, 0)); //dp[i][j]表示点燃所有下标(i, j)内的气球所能得到的最大钱数
        int n = len + 2;
        for (int k = 1; k <= n; k++) { //从长度为1的开始算起
            for (int left = 0; left < n - 1 - k; left++) {
                int right = left + k + 1;
                for (int i = left + 1; i < right; i++) {
                    dp[left][right] = max(dp[left][right], num[left] * num[i] * num[right] + dp[left][i] + dp[i][right]);
                }
            }
        }
        return dp[0][n - 1];
    }
};