1.直接递归超时 View Code 2、加入一个矩阵,依然超时 View Code 3.真正的动态规划 View Code
DP状态转移方法--leetcode115 技术标签: 算法 leetcode Leetcode115题: Given a string S and a string T, count the number of distinct subsequences of S which equals T. A subsequence of a string is a new string which is formed fr... 查看原文 [leetcode]792. Number of Matching Subsequences ...
解法2 (https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation) 求S[s..e] 的回文串数 当S[s] != S[e] 时,可得公式 dp[s][e] = dp[s+1][e] + dp[s][e-1] - dp[s+1][e-1]; 即(包含s + 两边都不含...
我们会用动态规划先求出包括空序列的所有子序列,再返回答案之前再减去空序列。 我们用 dp[k] 表示 S[0 .. k] 可以组成的不同子序列的数目。如果 S 中的所有字符都不相同,例如 S = "abcx",那么状态转移方程就是简单的 dp[k 27 11.3k 28 【微扰理论】动态规划+哈希表 从无重复字母的情况开始讨论 ...
Return true because “leetcode” can be segmented as “leet code”. 最先的想法就是brute force的一个一个试,用recursive的方法,但是速度太慢。Exceed Time Limit 1 2 3 4 5 6 7 8 9 10 11 12 13 14 public class Solution { public boolean wordBreak(String s, Set<String> dict) { if (s....
Best Time to Buy and Sell Stock III - LeetCodeleetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/ Say you have an array for which theith element is the price of a given stock on dayi. Design an algorithm to find the maximum profit. You may complete at mosttwo...
leetcode-1081. Smallest Subsequence of Distinct Characters -dict-and-stack -- -- 6:18 App leetcode-62. Unique Paths - DP 9 -- 10:26 App leetcode-2842 [Google] Count K-Subsequences of a String With Maximum Beauty 69 -- 10:33 App leetcode-210-course schedule II - Kahn's algorithm...
For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007. Sample Input 3 1 2 3 Sample Output 7 题意:找到一个字符串里面所有的不下降子序列的数量之和。
leetcode -- 115. Distinct Subsequences ;rabbbit”,T = "rabbit"为例: dp[i][j]表示T的从0开始长度为i的子串和S的从0开始长度为j的子串的匹配的个数。 比如, dp[2][3]表示T中的..."中的b2, 所以要把T中的"rab"和S中的"rab1"的匹配个数累加到当前的dp[3][4]中。 所以dp[3][4] +=...
https://leetcode.com/problems/distinct-subsequences/?tab=Description 一般没有明显思路的情况下,都要想想DP,用下Divide-and-Conque 下面的思路很好,很清晰 https://discuss.leetcode.com/topic/9488/easy-to-understand-dp-in-java/2 一个矩阵,如果字符不一样,那就用左侧的;如果字符一样,那就左侧+上方的。