PartyPartyAnimal的博客

tf.train.batch([example, label], batch_size=batch_size, capacity=capacity) [example, label]表示样本和样本标签 batch_size是返回的一个batch样本集的样本个数 capacity是队列中的容量。这主要是按顺序组合成一个batchtf.train.shuffle_batch([example, l

findKth(A,B,k)函数思路如下： 1. 保持A是短的那一个数组，B是长的 2. 平分k, 一半在A，一半在B （如果A的长度不足K/2,那就pa就指到最后一个） 3. 如果pa的值 < pb的值，那证明第K个数肯定不会出现在pa之前，递归，把A数组pa之前的砍掉，同理递归砍B数组。 4. 递归到 m == 0 （短的数组用完了） 就返回 B[k - 1], 或者k == 1（找第一

Given a string, find the length of the longest substring without repeating characters.Examples:Given “abcabcbb”, the answer is “abc”, which the length is 3.Given “bbbbb”, the answer is “b”, with the le

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it

Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the same ele

Python：class Solution(object): def isValid(self, s): """ :type s: str :rtype: bool """ stack = [None] #用来依次存储左括号 #用栈来操作

分析： 递归实现，如果左括号还有剩余，则可以放置左括号，如果右括号的剩余数大于左括号，则可以放置右括号。Python：class Solution(object): def generateParenthesis(self, n): """ :type n: int :rtype: List[str] """

思路： 先任意取三元素和res；记录和目标的差值diff 每次取一个数，在后续列表中找2个元素，满足三数之和最接近target，通过判断三数之和和目标的差值和diff大小，小于diff则将三数之和和差值赋值给res和diff,然后再分析和小于移动左指针，大于移动右指针，直到遍历找到最接近的数 Python：class Solution(object): def threeSumClose

python:class Solution(object): def threeSum(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ res = [] nums.sort() #先对数组

Write a function to find the longest common prefix string amongst an array of strings. Python：class Solution(object): def longestCommonPrefix(self, strs): """ :type strs: List[str]

class Solution(object): def myAtoi(self, str): """ :type str: str :rtype: int """ i = 0 sign = 1 base = 0 n = len(str) INT_