问题 给出字符串S和T,计算S中为T的不同的子序列的个数. 一个字符串的子序列是一个由该原始字符串通过删除一些字母(也可以不删)但是不改变剩下字母的相对顺序产生的一个新字符串.如,ACE是ABCDE的一个子序列,但是AEC不是. 这里有一个例子: S＝“rabbbit”,T＝“rabbit” 返回值应为3 初始思路 要找出子序列的个数,首先要有找出S中为T的子序列的方法.T是S的子序列,首先其每一个字母肯定会在S中出现,通过遍历T的每一个字母即可完成这个检查.而根据不能乱序的要求,下一个字母在S…

https://leetcode.com/problems/edit-distance/ Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Inse…

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 from the original string by deleting some (can be none) of the characters without disturbing the…

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…

［抄题］: 给出字符串S和字符串T,计算S的不同的子序列中T出现的个数. 子序列字符串是原始字符串通过删除一些(或零个)产生的一个新的字符串,并且对剩下的字符的相对位置没有影响.(比如,“ACE”是“ABCDE”的子序列字符串,而“AEC”不是). Here is an example:S = "rabbbit", T = "rabbit" Return 3. ［思维问题］: ［一句话思路］: 由于要查找T.最后一位相同时可以同时删,不相同时只能删S,不能多删除T.…

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 from the original string by deleting some (can be none) of the characters without disturbing the…

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 from the original string by deleting some (can be none) of the characters without disturbing the…

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…

引言 子序列和子字符串或者连续子集的不同之处在于,子序列不需要是原序列上连续的值. 对于子序列的题目,大多数需要用到DP的思想,因此,状态转移是关键. 这里摘录两个常见子序列问题及其解法. 例题1, 最长公共子序列 我们知道最长公共子串的求法,先温习一下,它的求法也是使用DP思想,对于 字符串s1 和字符串s2,令 m[i][j] 表示 s1上以s1[i]结尾的子串和s2上s2[j]结尾的子串的最长公共子串长度,因为公共子串必须是连续的,因此状态转移方程:m[i, j] = (s1[i] ==…

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/description/ 题目描述 Given a string S and a string T, count the number of distinct subsequences of S which equals…