Algorithm

每日一刷-Burst_Balloon

[每日一刷] (Burst Balloon)


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.

Example:
Given [3, 1, 5, 8]
Return 167

解题思路


看到maximum coins, 发现是一道求最优解的题，马上想到使用动态规划
本题的最优子结构是opt[i~j] = opt[i~k-1] + score[k] + opt[k+1~j], 其中k指的是最后一个爆破的气球的位置
本题的trick在于在算最优子结构的时候要在两边多算一个位置。例如算opt[1,2]的时候要将0, 3都算进去

代码虽然在leetcode上超时了，不过同样的做法使用JAVA和C++却可以通过，可能还是因为Python太慢了吧-.-

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#!/usr/bin/env 
# -*- coding:utf-8 -*-
"""
时间复杂度 O(n^3)
"""
prices = [1, 2, 3, 0, 2]
class Solution(object):
    def maxCoins(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        nums.insert(0, 1)
        nums.append(1)
        # table[i][j] stores the optimal result of bursting balloon starts at i and the length is j
        # balloon starts from 1 to n
        table = [[0 for i in xrange(n-I+1)] for I in xrange(n+1)]
        table.insert(0, []) 
        for i in xrange(1, n+1):
            table[i][1] = nums[i-1] * nums[i] * nums[i+1]
        for l in xrange(2, n+1): # length from 2 to n
            for j in xrange(1, n-l+2): # balloon can starts from balloon 1 to balloon n-l+1
                maximum = 0
                for i in xrange(j, j+l): # balloon i is the last burst balloon
                    if i < j+l-1:
                      temp = table[j][i-j] + table[i+1][j+l-i-1] + nums[i]*nums[j-1]*nums[j+l]
                    else:
                      temp = table[j][i-j] + nums[i]*nums[j-1]*nums[j+l]
                    maximum = max(maximum, temp)
                table[j][l] = maximum
        return table[1][n]

Algorithm

线段树-SegmentTree

线段树的优点:
  O(logn)时间内，在一段区间中求最大最小值

线段树的特征:

  • 二叉树

  • 每个节点表示一个区间的最大值

  • 左儿子 = 平分后左区间，右儿子 = 平分后右区间

  • 不可增减节点(线段数结点总数需要在最开始就给定)

  • 空间复杂度: O(n)

    • 线段树的操作:

    线段树的查询: O(logn)

  • 查询的区间 和 线段树节点区间 相等 -> 直接返回

  • 查询的区间 被 线段树节点区间 包含-> 递归向下搜索左右子树

  • 查询的区间 和 线段树节点区间 不相交 -> 结束

  • 查询的区间 和 线段树节点区间 相交且不相等 -> 分裂查询区间

    • 线段树的建立: O(n)

  • 自上而下递归分裂

  • 自下而上回溯更新

    • 线段树的更新: O(logn)

  • 自上而下递归查询

  • 自下而上回溯更新

    • 问题特征:

  • 对区间求sum

  • 对区间求Max/Min

  • 对区间求Count

    • 构建形式

    1. 下标作为建立区间
    2. 值作为建立区间

    Python:

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    #!/usr/bin/env 
    # -*- coding:utf-8 -*-
    class SegmentTree(object):
        def __init__(self, nums, s=None, e=None): # build
            """
            initialize your data structure here.
            :type nums: List[int]
            """
            n = len(nums)
            if s == None:
                self.start, self.end = 0, n-1
            else:
                self.start, self.end = s, e
            self.mid = self.start + (self.end-self.start)/2
            self.left, self.right = None, None
            self.val = 0
            if self.end < self.start:
                return
            if self.end == self.start:
                self.val = nums[self.start]
            else:
                self.left = SegmentTree(nums, self.start, self.mid)
                self.right = SegmentTree(nums, self.mid+1, self.end)
                self.val = self.left.val + self.right.val
            
        def update(self, i, val): # modify
            """
            :type i: int
            :type val: int
            :rtype: int
            """
            if i == self.start and i == self.end:
                self.val = val
            else:
                if i <= self.mid:
                    self.left.update(i, val)
                else:
                    self.right.update(i, val)
                self.val = self.left.val + self.right.val
            
        def sumRange(self, i, j): # query
            """
            sum of elements nums[i..j], inclusive.
            :type i: int
            :type j: int
            :rtype: int
            """
            if i == self.start and j == self.end: # equal
                return self.val
            elif self.start > j or self.end < i: # not intersect
                return 0
            else: # intersect
                if i > self.mid: # all at the right sub tree
                    return self.right.sumRange(i, j)
                elif j <= self.mid: # all at the left sub tree
                    return self.left.sumRange(i, j)
                else: # some at the right & some at the left
                    return self.left.sumRange(i, self.mid) + self.right.sumRange(self.mid+1, j)

    Algorithm

    每日一题-Best_Time_to_Buy_and_Sell_Stock_with_Cooldown

    [每日一题] (Best Time to Buy and Sell Stock with Cooldown)
    Say you have an array for which the ith element is the price of a given stock on day i.
    Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:
    You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
    After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

    解题思路


    本题使用动态规划，解题思路类似于抢劫问题。使用3个变量free, cool, have来存储当前位置的最优解。
    free用来存储到当前位置，状态为空闲时的最优解
    have用来存储到当前位置，状态为持有股票时的最优解
    cool用来存储到当前位置，状态为正在冷却时的最优解

    状态转移矩阵为：
    free = max(free, cool) # 当前free的最优解由上一时间free的最优解和cool的最优解得出
    have = max(have, free-p) # 当前have的最优解由上一时间持有股票时的最优解和上一时间空闲而这一时间购买股票的最优解得出
    cool = have + p #当前cool的最优解由上一时间持有股票的最优解再在这一时间卖出得出

    Python:

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    #!/usr/bin/env 
    # -*- coding:utf-8 -*-
    """
    时间复杂度 O(n)
    """
    prices = [1, 2, 3, 0, 2]
    class Solution(object):
      def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        free = 0
        have = cool = float('-inf')
        for p in prices:
            free, have, cool = max(free, cool), max(have, free-p), have+p
        return max(free, cool)