方法二. 修改状态定义(同时用到了贪心算法、二分查找)时间O(n * log n)空间 O(n) 参考:https://leetcode-cn.com/problems/longest-increasing-subsequence/solution/dong-tai-gui-hua-er-fen-cha-zhao-tan-xin-suan-fa-p/ 定义新状态(特别重要):tail[i]表示长度为i + 1的所有上升子序列的结尾的最小...
Code-It-Yourself! Tetris - Programming from Scratch (Quick and Simple C++) 0 0 15:31 App Maximum Product Subarray - Dynamic Programming - Leetcode 152 0 0 13:24 App Counting Bits - Dynamic Programming - Leetcode 338 - Python 0 0 18:25 App Longest Common Subsequence - Dynamic Programmin...
首先看长度等于1的情况: 看长度为2的情况: 看长度为3的情况: 重复上述步骤即可 结果图: 正文 问题描述 https://leetcode.com/problems/longest-palindromic-subsequence/description/ 一个例子: 令数组为a,画一个行列为a长度的矩阵,令为T 回到顶部 首先看长度等于1的情况: 也就是只看一个元素,比如只看a[0]...
[leetcode] 516. Longest Palindromic Subsequence Description Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000. Example 1: Input: "bbbab" Output: 4 One possible longest palindromic subsequence is "bbbb". 1. 2. 3...
【LeetCode】Longest Common Subsequence最长公共子序列(求出某一解+LCS长度) - Medium,LongestCommonSubsequence 给出两个字符串,找到最长公共子序列(LCS),返回LCS的长度。 说明 最长公共子序列的定义:
Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2 Output: 4 Explanation: The longest arithmetic subsequence is [7,5,3,1]. 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/longest-arithmetic-subsequence-of-given-difference 著作权归领扣网络所有。商业转载请联系官方授权,非商...
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. Note: There may be more than one LIS combination, it is only necessary for you to return the length. Your algorithm should run in O(n2) complexity. ...
tails[i] 表示, 长度为 i + 1的 increasing subsequence, 的最后一个数字。 所以如果 i + 1 > size, 那么我们需要扩大这个size Anyway, Good luck, Richardo! -- 09/26/2016 My code: publicclassSolution{publicintlengthOfLIS(int[]nums){if(nums==null||nums.length==0){return0;}List<Integer>li...
1. Description Longest Increasing Subsequence 2. Solution 解析:Version 1,最长递增子序列,典型的动态规划问题,定义状态:以nums[i]作为结尾元素的最长递增子序列的长度,状态转移方程:遍历nums[i]之前的元素nums[j],如果nums[i] > nums[j],则其最长递增子序列的长度为max(dp[i], dp[j] + 1),遍历之后,可...
Input: a = "aaa", b = "aaa" Output: -1 Explanation: Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a. So the answer would be -1.