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最新文章列表
Climbing Stairs
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 climb to the top?
爬楼梯的问题,思路和Unique Paths问题类似。我们假设到了 ...
Unique Paths II
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 is marked as 1 and 0 respectively in t ...
Unique Paths
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bot ...
Maximum Subarray
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has th ...
Wildcard Matching 通配符
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover th ...
Coin Change
Coin Change
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amoun ...
Wildcard Matching
Wildcard Matching
给定两个字符串p和q,判断两个字符串是否匹配。规则是:‘*’可以代表任意长度的字符串,也可以代表空串;‘?’代表任意单个的字符串。
例如:
isMatch("aa", "a") → false
isMatch("aa", "aa") → true
isMatch("aaa&q ...
时间序列分段算法 [Time series Breakout Detection]
在时间序列分析中,断点检测(breakout detection)是一个很基本的问题。
通过捕捉时序数据中的断点(breakout),来发现时序数据所表示的系统在过去是否发生了某种事件(event),进而为系统诊断提供必要的数据支持。
为了实现对时序断点的检测,我们首先需要对时序的整体时序做拟合。
这里我们通过一条直线来拟合一段时序,如果时序的趋势发生了变化,则用多条直线来拟合整条时 ...
LeetCode[动态规划] - #5 Longest Palindromic Substring
原题链接:#5 Longest Palindromic SubString
要求:
给定一个字符串S,找出它的最长回文子串。假定S的最大长度为1000,且最长回文子串唯一
难度:中等
分析:
假定字符串s为回文字符串,则在s头部和尾部分别添加相同字符串[x],所得结果s'=[x]s[x]也为回文字符串(论述1)。可使用动态规划方法解决此问题,递推公式便基于此特性。
创 ...
188 Best Time to Buy and Sell Stock IV——leetcode
class Solution {
public:
Solution(){}
int maxProfit(int K, vector<int> &prices) {
vector<int> lowVec;
vector<int > highVec;
if(prices ...