动态规划-Burst Balloons】的更多相关文章

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.…
参考:LeetCode 312. Burst Balloons(戳气球) java代码如下 class Solution { //参考:https://blog.csdn.net/jmspan/article/details/51208865 public int maxCoins(int[] nums) { int[] balls = new int[nums.length+2]; balls[0] = 1; balls[balls.length - 1] = 1; int[][] coins…
Burst Balloons (Medium) 这题没有做出来. 自己的思路停留在暴力的解法, 时间复杂度很高: 初始化maxCount = 0. 对于当前长度为k的数组nums, 从0到k - 1逐个选取第i个气球扎破 计算扎破气球得到的金币数, count = nums[i - 1] * nums[i] * nums[i + 1]. 从nums中删掉nums[i]. 查询m[nums]是否存在, 不存在则递归调用maxCoins(nums)并插入m[nums]. count += m[num…
lc 312 Burst Balloons 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] * num…
870. Advantage Shuffle 思路:A数组的最大值大于B的最大值,就拿这个A跟B比较:如果不大于,就拿最小值跟B比较 A可以改变顺序,但B的顺序不能改变,只能通过容器来获得由大到小的顺序,并且必须存储相应的index,因为最终需要将选择的A的数值存入与这个B相对应的index下 class Solution { public: vector<int> advantageCount(vector<int>& A, vector<int>&…
[LeetCode]452. Minimum Number of Arrows to Burst Balloons 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/ 题目描述: There are a number of spherical balloons spread in two-dimensio…
2018-10-03 19:29:43 问题描述: 问题求解: 很有意思的题目,首先想到的是暴力遍历解空间,当然也用到了memo,可惜还是TLE,因为时间复杂度确实有点过高了,应该是O(n!). Map<LinkedList, Integer> map = new HashMap<>(); public int maxCoins(int[] nums) { if (nums.length == 0) return 0; LinkedList<Integer> list…
这道题目做了两个晚上,发现解题思路的优化过程非常有代表性.文章详细说明了如何从回溯解法改造为分治解法,以及如何由分治解法过渡到动态规划解法.解法的用时从 超时 到 超过 95.6% 提交者,到超过 99.8% 提交者.现整理下来分享给大家,如有错误评论区欢迎指正! 题目如下: 回溯法 刚看到这个题目,脑中可以很轻易的想象出解空间的结构:一个n层的数组,每层的元素相同,我们从第一层走到第n层,每层走动时不能使用之前走过的元素.然后按照规则计算获取的金币,我们尝试所有可以走的路径并记录下每条路径所能…
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 r…
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 rig…