Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters....
我的LeetCode解题报告索引 题目链接 Given a stringSand a stringT, count the number of distinct subsequences ofTinS. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of ...
dp[j +1] += dp[j]returndp[-1]if__name__ =="__main__":assertSolution().numDistinct("rabbbit","rabbit") ==3 欢迎查看我的Github(https://github.com/gavinfish/LeetCode-Python) 来获得相关源代码。
115. Distinct Subsequences Given two strings s and t, returnthe number of distinctsubsequencesofswhich equalst. The test cases are generated so that the answer fits on a 32-bit signed integer. Example 1: Input:s = "rabbbit", t = "rabbit"Output:3Explanation:As shown below, there are 3 ...
Distinct Subsequences LeetCode:Distinct Subsequences 我的LeetCode解题报告索引 题目链接 Given a stringSand a stringT, count the number of distinct subsequences ofTinS. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the ...
leetcode:Distinct Subsequences lintcode:Distinct Subsequences Problem Statement Given a stringSand a stringT, count the number of distinct subsequences ofTinS. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters with...
【Leetcode】115. Distinct Subsequences 又是一道DP的题目,个人感觉DP的题目比较难,主要因为:(1)DP的难点是寻找子问题,如果找到很好的子问题,那么就可以瞬间搞定。(2)通常也会带有一点backtracking的思想,有时候总是优先想到bt,然后思路就被限制了。 DP在leetcode主要解决两类问题,第一是求数目,第二是求最优解...
解题方法 动态规划 日期 题目地址:https://leetcode.com/problems/distinct-subsequences/description/ 题目描述 Given a string S and a string T, count the number of distinct subsequences of S...
Return3. 原问题链接:https://leetcode.com/problems/distinct-subsequences/ 问题分析 这个问题相对来说比较难找到思路。对于给定的源串s来说,要让它的子串能够构成目标串t,那么理论上可能是它里面的任何一个元素可能存在或者删除。如果按照这个思路,假设源串长度为m的话,那么对它里面每种情况进行构造并和目标串比...
Distinct Subsequences@LeetCode Distinct Subsequences 动态规划题。先用二维动态规划的思路解释下:设match是动态规划表,其中match[i][j]表示S.substring(0, i)对T.substring(0, j)有几种组成方式,递推公式为: 若S.charAt(i - 1) == T.charAt(j - ),则match[[i][j] = match[i - 1][j - 1] ...