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.

Recurse:
Judge Small: Accepted!
Judge Large: Time Limit Exceeded

int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int slen = S.length();
int tlen = T.length();
if(slen <= tlen){
if(S == T) return 1;
else return 0;
} if(S[slen-1] != T[tlen-1]) return numDistinct(S.substr(0,slen-1), T);
else
return numDistinct(S.substr(0,slen-1), T) + numDistinct(S.substr(0,slen-1), T.substr(0,tlen-1));
}

dp:
Judge Small: Accepted!
Judge Large: Accepted!

int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int col = S.length() + 1;
int row = T.length() + 1;
int** dp = new int*[row];
for(int i = 0; i < row; ++i)
dp[i] = new int[col]; for(int i = 0; i < row; ++i)
dp[i][0] = 0;
for(int j = 0; j < col; ++j)
dp[0][j] = 1; for(int i = 1; i < row; ++i)
for(int j = 1; j < col; ++j)
if(T[i-1] == S[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i][j-1];
else dp[i][j] = dp[i][j-1]; int tmp = dp[row-1][col-1]; for(int i = 0; i < row; ++i)
delete[] dp[i];
delete[] dp; return tmp;
}

leetcode_question_115 Distinct Subsequences的更多相关文章

  1. [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 ...

  2. Distinct Subsequences

    https://leetcode.com/problems/distinct-subsequences/ Given a string S and a string T, count the numb ...

  3. Leetcode Distinct Subsequences

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  4. 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 ...

  5. [Leetcode][JAVA] Distinct Subsequences

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  6. Distinct Subsequences Leetcode

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  7. 【leetcode】Distinct Subsequences(hard)

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  8. 【LeetCode OJ】Distinct Subsequences

    Problem Link: http://oj.leetcode.com/problems/distinct-subsequences/ A classic problem using Dynamic ...

  9. 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  ...

随机推荐

  1. jcSQL简明执行流程图

    赶着"黑色七月"的最后一天发一篇记点东西,这个月一共掉了三架飞机,我一直很害怕坐着一架人造的东西飞在几万米的高空,相比自己长出一对翅膀,前者应该要脆弱很多.这些人每个人都因为不同的 ...

  2. MySQLdb-python的安装

    第一步下载: 第一步:进入https://github.com/farcepest/MySQLdb1/ 第二步:解压 Shell>unzip /root/MySQLdb1-MySQLdb-1.3 ...

  3. SQL Server 分离与附加数据库

    前期准备: 创建数据库 create database Studio     on primary     (name ='Studio',filename ='E:\Studio.mdf'),    ...

  4. 好的组件,无须太复杂 – KISSY Slide 组件简介

    KISSY Slide 组件首页:http://gallery.kissyui.com/slide/1.1/guide/index.html V1.1 New Featurs Slide是一个幻灯切换 ...

  5. Android Studio优化之启用Shift+Ctrl+O导入所有的包

    在使用Eclipse开发Android应用时,开发者往往会使用Shift+Ctrl+O快捷键来快速导入所有的包,和移除已经导入但还未使用的包.但这个快捷键在Android Studio没人是给有开启的 ...

  6. Unix/Linux环境C编程入门教程(20) 搭建基于Mac的 Xcode 与 QT 开发环境

    1.启动 Vmware,如果没有 VMware 的同学,请看前面我们搭建 VMware 的视频 2.打开虚拟机以后,出现虚拟机界面 3 新建一个虚拟机 4 选择自定义,单击下一步 5 选择默认的 VM ...

  7. 【POJ】1330 Nearest Common Ancestors ——最近公共祖先(LCA)

    Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18136   Accept ...

  8. 剑指offer32 整数中1出现的次数(从1到n整数中1出现的次数)

    class Solution { public: int NumberOf1Between1AndN_Solution(int n) { ) ; ; ; while(i<=n) { int p= ...

  9. iOS开发之GCD使用总结

    GCD是iOS的一种底层多线程机制,今天总结一下GCD的常用API和概念,希望对大家的学习起到帮助作用. GCD队列的概念 在多线程开发当中,程序员只要将想做的事情定义好,并追加到DispatchQu ...

  10. [转载]解析WINDOWS中的DLL文件---经典DLL解读

    [转载]解析WINDOWS中的DLL文件---经典DLL解读 在Windows世界中,有无数块活动的大陆,它们都有一个共同的名字——动态链接库.现在就走进这些神奇的活动大陆,找出它们隐藏已久的秘密吧! ...