用dp解 1)up定义为nums[i-1] < nums[i] down nums[i-1] > nums[i] 两个dp数组, up[i],记录包含nums[i]且nums[i-1] < nums[i]的最长子序列长度 down[], 记录包含nums[i]nums[i-1] > nums[i]的最长子序列长度 2)更新方程 有三种情况 nums[i-1] <or=or> nums[i] a) up[i] = down[i-1] + 1; down[i] = down…

Given a string s and a string t, check if s is subsequence of t. You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100). A subsequence of…

A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two…

Another interesting DP. Lesson learnt: how you define state is crucial.. 1. if DP[i] is defined as, longest wiggle(up\down) subseq AT number i, you will have O(n^2) solution class Solution { struct Rec { Rec(): mlen_dw(), mlen_up(){} Rec(int ldw, int…

前言 这道题的实现方法有很多,包括dp,贪心算法,二分搜索,普通实现等等. 题目 Given a string s and a string t, check if s is subsequence of t. You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is…

Question Given a string s and a string t, check if s is subsequence of t. You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100). A subseq…

There are 3 possible approaches: DP, divide&conquer and greedy. And apparently, DP has O(n^2) complexity (TLE), DivideConquer can only be worse. Greedy can be O(n)! And its code is amazingly brief: class Solution { public: bool isSubsequence(string s…

A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, these are arithmetic sequences: 1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9 The follo…

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…

Given an unsorted array of integers, find the length of longest increasing subsequence. For example, Given [10, 9, 2, 5, 3, 7, 101, 18], The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than…