【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 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：对于该题，我们可以逆向思考，也就是无论我们前面如何选择气球，最后选择的总是剩下的最后那一个气球，再将那个气球与两个1相乘。定义一函数res获取从0 -> n-1 下标序列气球的最大收益值，则收益可表示为：从中选取一个气球作为最后爆炸的气球i，则收益可表示为：res(0,i-1) + 1*nums[i] * i+ res(i + 1,n-1);并采用迭代的方式记录下最大值，并将每种序列的最大值均记录在二维数组中，避免重复计算.

具体实现代码如下：

class Solution {
    public int maxCoins(int[] nums) {
        int[][] dp = new int[nums.length][nums.length];
        return maxCoins(nums,dp,0,nums.length - 1);
    }
    
    public int get(int[] nums,int i) {
        if(i==-1 || i==nums.length) {
            return 1;
        }
        return nums[i];
    }
    
    public int maxCoins(int[] nums,int[][] dp,int start,int end){
        if(start > end) {
            return 0;
        }
        if(dp[start][end] != 0) {
            return dp[start][end];
        }
        
        int max = nums[start];
        for(int i=start;i<=end;i++) {
            int val = maxCoins(nums,dp,start,i - 1) + 
                get(nums,start - 1) * get(nums,i) * get(nums,end + 1) +
                maxCoins(nums,dp,i + 1,end);
            max = Math.max(max,val);
        }
        dp[start][end] = max;//记录
        
        return max;
    }
}

leetcode上beat 100 %的代码

class Solution {
    /**
     * TIme: O(n^3); Space: O(n^2)
     * Imp points:
     * 1. dp[i][j] = for k in i, j: max(dp[i][k-1] + dp[k+1][j] + nums[k]*nums[i-1]*nums[j+1])
     */
    public int maxCoins(int[] nums) {
        int[] balloons = new int[nums.length+2];
        int n = 1;
        balloons[0] = 1;
        for (int num : nums) {
            if (num > 0) {
                balloons[n++] = num;
            }
        }
        balloons[n++] = 1;
        
        int[][] dp = new int[n][n];
        return maxCoins(0, n-1, balloons, dp);
    }
    
    private int maxCoins(int left, int right, int[] balloons, int[][] dp){
        if (right - left <= 1){
            return 0;
        }
        
        int max = 0;
        for (int k = left+1; k < right; k++){
            int maxLeft = dp[left][k] > 0 ? dp[left][k] : maxCoins(left, k, balloons, dp);
            int maxRight = dp[k][right] > 0 ? dp[k][right] : maxCoins(k, right, balloons, dp);
            max = Math.max(max, maxLeft + maxRight + balloons[k]*balloons[left]*balloons[right]);
        }
        dp[left][right] = max;
        return max;
    }
}