每周LeetCode算法题（十六）312. Burst Balloons

每周LeetCode算法题（十六）

题目: 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.

解法分析

本题需要用O(n^3)的复杂度。直观上，三层循环，一层窗口长度，由小到大；一层窗口的左右端点；一层窗口内每个位置与上次窗口长度的关系，为dp[left][right] = max(dp[left][right], dp[left][i] + dp[i][right] + num[left] * num[i] * num[right])，其中，dp的两维分别表示窗口的左右端点下标。

解释是，第三层循环选择窗口内的一个气球扎破，第二层和第一层循环选择窗口位置和大小。因为窗口由小变到大，大窗口内情况已经由小窗口决定。

C++代码

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int len = nums.size();
        int * num = new int[len + 2];
        for (int i = 1; i < len + 1; i++) {
            num[i] = nums[i - 1];
        }
        num[0] = 1;
        num[len + 1] = 1;

        int **dp = new int*[len + 2];
        for (int i = 0; i < len + 2; i++) {
            dp[i] = new int[len + 2];
            memset(dp[i], 0, sizeof(int) * (len + 2));
        }
        for (int width = 3; width <= len + 2; width++) {
            for (int left = 0; left + width - 1 <= len + 1; left++) {
                int right = left + width - 1;
                for (int i = left + 1; i < right; i++) {
                    dp[left][right] = max(dp[left][right], dp[left][i] + dp[i][right] + num[left] * num[i] * num[right]);
                }
            }
        }
        return dp[0][len + 1];
    }
};