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09 2017 档案

摘要:2分类1隐层nn, 作业默认设置: 1个输出单元, sigmoid激活函数. (因为二分类); 4个隐层单元, tanh激活函数. (除作为输出单元且为二分类任务外, 几乎不选用 sigmoid 做激活函数); n_x个输入单元, n_x为训练数据维度; 总的来说共三层: 输入层(n_x = X.s 阅读全文
posted @ 2017-09-27 10:26 英雄与侠义的化身 阅读(563) 评论(0) 推荐(0) 编辑

摘要:向量化地实现 Logistic Regression 算法. 阅读全文
posted @ 2017-09-25 08:37 英雄与侠义的化身 阅读(285) 评论(0) 推荐(0) 编辑

摘要:Find the contiguous subarray within an array (containing at least one number) which has the largest product. For example, given the array , the contig 阅读全文
posted @ 2017-09-16 12:38 英雄与侠义的化身 阅读(133) 评论(0) 推荐(0) 编辑

摘要:Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be 阅读全文
posted @ 2017-09-13 11:10 英雄与侠义的化身 阅读(77) 评论(0) 推荐(0) 编辑

摘要:Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. For example, If n = 4 and k = 2, a solution is: 做了好几个这种题了,就是 阅读全文
posted @ 2017-09-12 16:55 英雄与侠义的化身 阅读(116) 评论(0) 推荐(0) 编辑

摘要:Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, have the following unique permutati 阅读全文
posted @ 2017-09-12 12:04 英雄与侠义的化身 阅读(170) 评论(0) 推荐(0) 编辑

摘要:vector 判断是否存在某元素: 阅读全文
posted @ 2017-09-11 16:50 英雄与侠义的化身 阅读(166) 评论(0) 推荐(0) 编辑

摘要:Given a collection of distinct numbers, return all possible permutations. For example, [1,2,3] have the following permutations: 做了几个 backtrack 类的题目了, 阅读全文
posted @ 2017-09-11 16:24 英雄与侠义的化身 阅读(146) 评论(0) 推荐(0) 编辑

摘要:Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each num 阅读全文
posted @ 2017-09-11 11:58 英雄与侠义的化身 阅读(297) 评论(0) 推荐(0) 编辑

摘要:Given a set of candidate numbers (C) ( without duplicates ) and a target number (T), find all unique combinations in C where the candidate numbers sum 阅读全文
posted @ 2017-09-11 09:37 英雄与侠义的化身 阅读(124) 评论(0) 推荐(0) 编辑

摘要:vector 弹出末尾元素 v.pop_back(); 阅读全文
posted @ 2017-09-11 09:11 英雄与侠义的化身 阅读(735) 评论(0) 推荐(0) 编辑

摘要:Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possib 阅读全文
posted @ 2017-09-10 12:48 英雄与侠义的化身 阅读(411) 评论(0) 推荐(0) 编辑

摘要:Deap Learning(Ng) 学习笔记 author: 相忠良(Zhong Liang Xiang) start from: Sep. 8st, 2017 1 深度学习概论 打字太麻烦了,索性在吴老师的 text note 上直接标注,写出自己的总结和心得. 每一节,我都会用数字标出核心内容, 阅读全文
posted @ 2017-09-08 12:03 英雄与侠义的化身 阅读(306) 评论(0) 推荐(0) 编辑

摘要:Given a string s consists of upper/lower case alphabets and empty space characters , return the length of last word in the string. If the last word do 阅读全文
posted @ 2017-09-07 21:47 英雄与侠义的化身 阅读(137) 评论(0) 推荐(0) 编辑

摘要:知道啥是 gray code 就是收获了. 下面介绍了 gray code 发明的 motivation, 了解动机后就知道啥是 gray code 了. https://zh.wikipedia.org/wiki/格雷码 阅读全文
posted @ 2017-09-07 17:25 英雄与侠义的化身 阅读(190) 评论(0) 推荐(0) 编辑

摘要:Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, 阅读全文
posted @ 2017-09-07 16:17 英雄与侠义的化身 阅读(119) 评论(0) 推荐(0) 编辑

摘要:Given a collection of integers that might contain duplicates, nums , return all possible subsets. Note: The solution set must not contain duplicate su 阅读全文
posted @ 2017-09-06 17:29 英雄与侠义的化身 阅读(112) 评论(0) 推荐(0) 编辑

摘要:1. 声明一个 的二维数组: 2. 这么写结果不对: 我晕! 3. 不能妄图写成这样: 有点边迭代,边修改数组的意思. 4. push_back() 和 back() 的区别: coll.push_back()是把一个元素,放入这个容器的末尾,相当于末尾添加一个元素; coll.back()是获取最 阅读全文
posted @ 2017-09-06 14:57 英雄与侠义的化身 阅读(744) 评论(0) 推荐(0) 编辑

摘要:Given a set of distinct integers, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If , a so 阅读全文
posted @ 2017-09-06 12:08 英雄与侠义的化身 阅读(252) 评论(0) 推荐(0) 编辑

摘要:Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step .) You have 阅读全文
posted @ 2017-09-05 16:40 英雄与侠义的化身 阅读(119) 评论(0) 推荐(0) 编辑

摘要:Given an m x n grid filled with nonnegative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. 阅读全文
posted @ 2017-09-02 11:59 英雄与侠义的化身 阅读(123) 评论(0) 推荐(0) 编辑

摘要:You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you cl 阅读全文
posted @ 2017-09-01 11:46 英雄与侠义的化身 阅读(122) 评论(0) 推荐(0) 编辑

摘要:Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space 阅读全文
posted @ 2017-09-01 11:15 英雄与侠义的化身 阅读(114) 评论(0) 推荐(0) 编辑

摘要:声明3行4列的数组 cpp const int m = 3, n = 4; vector A(m); // 3 rows for(int i = 0; i 阅读全文
posted @ 2017-09-01 09:34 英雄与侠义的化身 阅读(213) 评论(0) 推荐(0) 编辑

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