算法导论的第四章对于divide-conquer进行了阐述, 感觉这本书特别在,实际给出的例子并不多,更多其实是一些偏向数学性质的分析, 最重要的是告诉你该类算法分析的一般性策略. 估计 首先是估计算法的时间复杂度,这里我感觉大多数情况下该类算法的时间复杂度可以由两种策略来完成. master method 这种方式简单, 准确, 个人认为一般能用这种尽量使用这种. 对于常数 a >= 1, b > 1, T(n) = a T ( n / b ) + f(n), 也就是说算法T对于规模为n的问…
分治法基础 分治法(Divide and Conquer)顾名思义,思想核心是将问题拆分为子问题,对子问题求解.最终合并结果,分治法用伪代码表示如下: function f(input x size n) if(n < k) solve x directly and return else divide x into a subproblems of size n/b call f recursively to solve each subproblem Combine the results…
1.Implement exercise 2.3-7. 2. Implement priority queue. 3. Implement Quicksort and answer the following questions. (1) How many comparisons will Quicksort do on a list of n elements that all have the same value? (2) What are the maximum and minimum…
链接:https://leetcode.com/tag/divide-and-conquer/ [4]Median of Two Sorted Arrays [23]Merge k Sorted Lists [53]Maximum Subarray (2019年1月23日, 谷歌tag复习) 最大子段和. 题解: follow up 是divide and conquer If you have figured out the O(n) solution, try coding another…
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q…
Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and doe…
Moo University - Financial Aid 其实是老题了http://www.cnblogs.com/Philip-Tell-Truth/p/4926008.html 这一次我们换二分法来做这一道题,其实二分法比我以前那个方法好想一点,主要是这次我们可以根据下标进行二分,然后排两次序,第一次是根据分数来排序,然后记录分数的序,接下来就把aid排一次序,然后我们就可以进行二分了 参考http://www.hankcs.com/program/cpp/poj-2010-moo-un…