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:
- 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
思路：從整個過程來看，最後一步總是剩下一個數字來和兩個1相乘。也就是說，這個數字兩邊的區域的數字並沒有混乘故這個最後的數字可以把整個序列劃分爲兩個部分，這就是動態規劃結構的構造基礎。其狀態轉移方程爲：
dp[left][right] = max(dp[left][right], nums[left-1]*nums[k]*nums[right+1] + dp[left][k-1] + dp[k+1][right]);
對於[3, 1, 5, 8], dp數組的第一次循環結果更新了dp[1][1]、dp[2][2]、dp[3][3]、dp[4][4]，由於下標left<=right,因而dp矩陣爲上三角型矩陣。
0 0 0 0 0 0
0 3 0 0 0 0
0 0 15 0 0 0
0 0 0 40 0 0
0 0 0 0 40 0
0 0 0 0 0 0
第二次循環更新了dp[1][2]、dp[2][3]、dp[3][4]:
0 0 0 0 0 0
0 3 30 0 0 0
0 0 15 135 0 0
0 0 0 40 48 0
0 0 0 0 40 0
0 0 0 0 0 0
第三次循環更新了dp[1][3]、dp[2][4] :
0 0 0 0 0 0
0 3 30 159 0 0
0 0 15 135 159 0
0 0 0 40 48 0
0 0 0 0 40 0
0 0 0 0 0 0
第四次更新了dp[1][4] (最終結果):
0 0 0 0 0 0
0 3 30 159 167 0
0 0 15 135 159 0
0 0 0 40 48 0
0 0 0 0 40 0
0 0 0 0 0 0
代碼如下：
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) //len爲區間長度
for (int left = 1; left <= n - len + 1; ++left) { //k的左邊界
int right = left + len - 1; //k的右邊界
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];
}
};