题目如下: Given a string S, count the number of distinct, non-empty subsequences of S . Since the result may be large, return the answer modulo 10^9 + 7. Example 1: Input: "abc" Output: 7 Explanation: The 7 distinct subsequences are "a", &…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 动态规划 日期 题目地址:https://leetcode.com/problems/distinct-subsequences-ii/description/ 题目描述 Given a string S, count the number of distinct, non-empty subsequences of S . Since the re…

Given a string S, count the number of distinct, non-empty subsequences of S . Since the result may be large, return the answer modulo 10^9 + 7. Example 1: Input: "abc" Output: 7 Explanation: The 7 distinct subsequences are "a", "b…

Given a string S, count the number of distinct, non-empty subsequences of S . Since the result may be large, return the answer modulo 10^9 + 7. Example 1: Input: "abc" Output: 7 Explanation: The 7 distinct subsequences are "a", "b…

题目链接:https://leetcode-cn.com/problems/distinct-subsequences/description/ 参考链接:https://www.cnblogs.com/springfor/p/3896152.html http://blog.csdn.net/abcbc/article/details/8978146 dp[i][j]:S使用前i个字符,T使用前面j个字符.dp[0][0]使用S前0个字符,使用T前0个字符. 当T为空的时候,空字符串是任何S串…

2020-02-06 17:01:36 问题描述: 问题求解: 非常经典的计数dp问题,思路就是统计以每个字符为结尾的个数,最后求和即可. dp[i] = sum of (dp[j]) 0 <= j <= i:可以理解为将最后的一个字符追加到前面的字符串后面. 问题是如何去重. 当我们遇到相同的字符的时候,首先最后一个字符单独最为subseq要删除,因为前面计算过了,其次,只能加到第一次碰到形同字符的位置,因为再前面的在这个重复字符的位置已经计算过了. int mod = (int)1e9 +…

Distinct Subsequences 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 di…

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…

题目连接:10069 - Distinct Subsequences 题目大意:给出两个字符串x (lenth < 10000), z (lenth < 100), 求在x中有多少个z. 解题思路:二维数组DP, 有类似于求解最长公共子序列, cnt[i][j]表示在x的前j个字符中有多少个z 前i个字符. 状态转移方程 1.x[j] != z[i]              cnt[i][j] = cnt[i][j - 1]; 2.x[j] == z[i]   cnt[i][j] = cnt…

题目链接 题目要求: 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 th…