Distinct Subsequences
https://leetcode.com/problems/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 disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
定义f[i][j]表示在S[0,i]中,T[0,j]出现了几次。无论s[i]和t[j]是否相等,如果不匹配s[i],则f[i][j]=f[i-1][j];若s[i]==s[j],
f[i][j]=f[i-1][j]+[i-1][j-1]。
另外,当t=""时,只有一种匹配方式,f[i][0]=1;当s="",t!=""时,无论如何无法匹配,此时f[0][j]=0。
参考:http://www.cnblogs.com/yuzhangcmu/p/4196373.html
int numDistinct(string s, string t) {
int m=s.size();
int n=t.size();
vector<vector<int>> f(m+,vector<int>(n+,));
for(int i=;i<=m;i++)
{
for(int j=;j<=n;j++)
{
if(i== && j==)
f[i][j]=;
else if(i==)
f[i][j]=;
else if(j==)
f[i][j]=;
else
f[i][j]=f[i-][j]+(s[i-]==t[j-]?f[i-][j-]:);
}
}
return f[m][n];
}
Distinct Subsequences的更多相关文章
- [LeetCode] Distinct Subsequences 不同的子序列
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- Leetcode Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- LeetCode(115) Distinct Subsequences
题目 Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequen ...
- [Leetcode][JAVA] Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- Distinct Subsequences Leetcode
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- 【leetcode】Distinct Subsequences(hard)
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- 【LeetCode OJ】Distinct Subsequences
Problem Link: http://oj.leetcode.com/problems/distinct-subsequences/ A classic problem using Dynamic ...
- LeetCode 笔记22 Distinct Subsequences 动态规划需要冷静
Distinct Subsequences Given a string S and a string T, count the number of distinct subsequences of ...
- leetcode 115 Distinct Subsequences ----- java
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
随机推荐
- 新建structs2 web应用及structs.xml常用基础配置
建立一个structs2 web应用程序 1. 创建一个基本的web应用程序 2. 添加structs2的jar文件到Class Path 将structs2的最小jar包拷到WEB-INF/lib目 ...
- [LeetCode] Merge Sorted Array 混合插入有序数组
Given two sorted integer arrays A and B, merge B into A as one sorted array. Note:You may assume tha ...
- [LeetCode] Two Sum 两数之和
Given an array of integers, return indices of the two numbers such that they add up to a specific ta ...
- C#扩展方法
扩展方法使您能够向现有类型“添加”方法,而无需创建新的派生类型.重新编译或以其他方式修改原始类型. 扩展方法就相当于一个马甲,给一个现有类套上,就可以为这个类添加其他方法了. 马甲必须定义为stati ...
- 字节、字、bit、byte的关系
字 word 字节 byte 位 bit 字长是指字的长度 1字=2字节(1 word = 2 byte) 1字节=8位(1 byte = 8bit) 一个字的字长为16 一个字节的字长是8 bps ...
- 修改hosts文件在本地使域名解析到指定IP
# Additionally, comments (such as these) may be inserted on individual # lines or following the mac ...
- Android Studio中的CmakeList NDK配置
Android Studio2.2之后直接可以在创建工程时添加NDK支持了,添加之后,main文件夹下会多出一个native-lib.cpp这个文件,如果只为了一个简单的NDK接口,貌似这就结束了.直 ...
- bzoj3052: [wc2013]糖果公园
又是一代神题. uoj测速rank10,bzoj测速rank26(截止当前2016.5.30 12:58) 带修改的树上莫队. 修改很少,块的大小随便定都能A 然而我一开始把开3次根写成了pow(bl ...
- 基于Dubbo框架构建分布式服务(一)
Dubbo是Alibaba开源的分布式服务框架,我们可以非常容易地通过Dubbo来构建分布式服务,并根据自己实际业务应用场景来选择合适的集群容错模式,这个对于很多应用都是迫切希望的,只需要通过简单的配 ...
- 脚手架搭建的vue项目里引入jquery和bootstrap
引入jquery: 1.在cmd输入:npm install jquery,回车,等待.. 2.在webpack.base.conf.js里进行如下操作: 3.在webpack.prod.conf.j ...