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)

    Algorithm

    每日一题-A+B-

    [每日一题] (A+B-)
    给出两个字符串,s1和s2。s2中的+表示前面的字符重复2次，-代表前面的字符重复4次。问题是求s1中有多少个连续或不连续的s2。
    例子: s1 = “aaabbbb” , s2 = “a+b-“
    输出: 3

    解题思路


    使用动态规划。具体思路说起来有点绕…总之算是一个比较麻烦的动态规划题。

    用n数组存储到当前位置为止有多少个n子串，用n_1数组存储到当前位置为止有多少个n-1子串
    状态转移函数是n[i] = n[i-1] + n_1[i-1] + n_1[i-2] + … + n_1[2]或n_1[4]，大概就是这个意思

    代码没有经过很多测试，可能会有bug

    Python:

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    #!/usr/bin/env python
    # -*- coding:utf-8 -*-
    #s1 = "aabbbbaacc"
    #s2 = "a+b-c+"
    #s1 = "aaabababbaabbccc"
    #s2 = "a+b-c+"
    #s1 = "aababaaabaaab"
    #s1 = "aabaabaabcc"
    #s2 = "a+b+c+"
    #s1 = "waeginsapnaabangpisebbasepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc"
    #s2 = "a+b+c-"
    s1 = "wweqwqaabddakswbbcc"
    s2 = "a+b+"
    class Solution(object):
        def aPlusbPlus(self, s1, s2):
            pre = {} # 记录当前字符的前一字符
            index = {} # 记录当前字符位置
            count = {} # 记录当前字符找了几个
            bound = {} # 记录当前字符要找几个
            p = None
            for i in xrange(len(s2)/2):
                pre[s2[i*2]] = p
                index[s2[i*2]] = 0
                count[s2[i*2]] = 0
                bound[s2[i*2]] = 2 if s2[i*2+1] == '+' else 4
                p = s2[i*2]
            n_1 = [[0, 0] for i in xrange(len(s1))]
            n = [[0, 0] for i in xrange(len(s1))]
            for i in xrange(len(s1)):
                c = s1[i]
                if c in index.keys():
                    if c == s2[0]: # 当前字符为s2首字符
                        count[c] += 1
                        sum = 1
                        for j in xrange(bound[c]):
                            sum *= count[c] - j
                        n[i] = [sum / bound[c], c]
                        n_1[i] = [0, c]
                        index[c] = i
                    else:
                        count[c] += 1
                        n_1[i] = [n[index[pre[c]]][0], c]
                        if count[c] < bound[c]:
                            n[i] = [0, c]
                        else :
                            n[i] = [n[index[c]][0], c]
                            count1, j = count[c] - 1, i-1
                            while count1 >= bound[c]-1 and j >= 0:
                                if n_1[j][1] == c:
                                    n[i][0] += n_1[j][0]
                                    count1 -= 1
                                j -= 1
                        index[c] = i
            return n[index[s2[len(s2)-2]]][0]