LeetCode 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 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

自己的解法（DP+分治法）：

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        if(n == 0) return 0;
        if(n == 1) return nums[0];
        vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
        //dp[i][j] begin from i, length is j
        dp[0][1] = nums[0] * nums[1];
        dp[n-1][1] = nums[n-1] * nums[n-2];
        for(int i = 1; i<n-1; i++)
            dp[i][1] = nums[i-1] * nums[i] * nums[i+1];
        for(int j = 2; j<=n; j++){
            for(int i = 0; i<=n-j; i++){
                for(int k = 0; k<j; k++){
                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[i+k+1][j-k-1] + nums[i+k] * (i>=1 ? nums[i-1]: 1) * (i+j<n ? nums[i+j]: 1));
                }
            }
        }
        return dp[0][n];
    }
};

答案中优秀的解法：

int maxCoins(vector<int>& nums) {
    int N = nums.size();
    nums.insert(nums.begin(), 1);
    nums.insert(nums.end(), 1);

    // rangeValues[i][j] is the maximum # of coins that can be obtained
    // by popping balloons only in the range [i,j]
    vector<vector<int>> rangeValues(nums.size(), vector<int>(nums.size(), 0));
    
    // build up from shorter ranges to longer ranges
    for (int len = 1; len <= N; ++len) {
        for (int start = 1; start <= N - len + 1; ++start) {
            int end = start + len - 1;
            // calculate the max # of coins that can be obtained by
            // popping balloons only in the range [start,end].
            // consider all possible choices of final balloon to pop
            int bestCoins = 0;
            for (int final = start; final <= end; ++final) {
                int coins = rangeValues[start][final-1] + rangeValues[final+1][end]; // coins from popping subranges
                coins += nums[start-1] * nums[final] * nums[end+1]; // coins from final pop
                if (coins > bestCoins) bestCoins = coins;
            }
            rangeValues[start][end] = bestCoins;
        }
    }
    return rangeValues[1][N];
}