算法:Common Subsequence(动态规划 Java 最长子序列)
Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
0
求最大子序列的问题:
递推关系:c[][]存储最长子序列长度
AC代码:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str1, str2;
while (sc.hasNext()){
str1 = sc.next();
str2 = sc.next();
System.out.println(LcsLength(str1.toCharArray(), str2.toCharArray()));
}
sc.close();
}
public static int LcsLength(char a[], char b[]){
int aL = a.length;
int bL = b.length;
int ans[][] = new int[aL + 1][bL + 1];
for (int i = 0; i <= aL; i++) {
ans[i][0] = 0;
}
for (int i = 0; i <= bL; i++) {
ans[0][i] = 0;
}
for (int i = 1; i <= aL; i++) {
for (int j = 1; j <= bL; j++) {
if (a[i - 1] == b[j - 1])
ans[i][j] = ans[i - 1][j - 1] + 1;
else if (ans[i][j - 1] >= ans[i - 1][j])
ans[i][j] = ans[i][j - 1];
else
ans[i][j] = ans[i - 1][j];
}
}
return ans[aL][bL];
}
}
可以输出最长子序列之一的代码:
import java.util.Scanner;
public class Main {
public static int x[][];
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str1, str2;
while (sc.hasNext()) {
str1 = sc.next();
str2 = sc.next();
System.out.println("最大子序列长度: " + LcsLength(str1.toCharArray(), str2.toCharArray()));
}
sc.close();
}
public static int LcsLength(char a[], char b[]) {
int aL = a.length; // a数组的长度
int bL = b.length; // b数组的长度
int ans[][] = new int[aL + 1][bL + 1]; // 存储i,j当前的最长子序列长度
x = new int[aL + 1][bL + 1]; // 存储ans[i][j]是斜对角还是左边,上边来的
for (int i = 0; i <= aL; i++) {
ans[i][0] = 0; // 空的子序列都为0
}
for (int i = 0; i <= bL; i++) {
ans[0][i] = 0; // 空的子序列都为0
}
for (int i = 1; i <= aL; i++) {
for (int j = 1; j <= bL; j++) {
if (a[i - 1] == b[j - 1]) { // 判断字符是否相等
ans[i][j] = ans[i - 1][j - 1] + 1; // 相等长度加1
x[i][j] = 1; // 标记为相等斜着来的
} else if (ans[i][j - 1] >= ans[i - 1][j]) { // 左边大于上边就用最大的
ans[i][j] = ans[i][j - 1];
x[i][j] = 2; // 标记为左边来的
} else {
ans[i][j] = ans[i - 1][j];
x[i][j] = 3; // 标记为上边来的
}
}
}
System.out.print("最大子序列: ");
Lcs(aL, bL, a);
System.out.print('\n');
return ans[aL][bL];
}
// 递归输出最大子序列其中一个
public static void Lcs(int i, int j, char a[]) {
if (i == 0 || j == 0)
return;
if (x[i][j] == 1) {
Lcs(i - 1, j - 1, a);
System.out.print(a[i - 1]);
} else if (x[i][j] == 2)
Lcs(i, j - 1, a);
else if (x[i][j] == 3)
Lcs(i - 1, j, a);
}
}
算法:Common Subsequence(动态规划 Java 最长子序列)的更多相关文章
- HDU 1159 Common Subsequence 动态规划
2017-08-06 15:41:04 writer:pprp 刚开始学dp,集训的讲的很难,但是还是得自己看,从简单到难,慢慢来(如果哪里有错误欢迎各位大佬指正) 题意如下: 给两个字符串,找到其中 ...
- hdu1422重温世界杯(动态规划,最长子序列)
重温世界杯 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submi ...
- nyoj 17-单调递增最长子序列 && poj 2533(动态规划,演算法)
17-单调递增最长子序列 内存限制:64MB 时间限制:3000ms Special Judge: No accepted:21 submit:49 题目描述: 求一个字符串的最长递增子序列的长度 如 ...
- 动态规划求最长公共子序列(Longest Common Subsequence, LCS)
1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与 ...
- hdu 1159:Common Subsequence(动态规划)
Common Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- HDU 1159 Common Subsequence (动态规划、最长公共子序列)
Common Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- 算法-最长子序列和C/C++实现(三个复杂度)
最长子序列和的问题非常easy: 就是一个数组,求出当中当中连续的某一段和,而这一段和是全部的连续段和的最大的值.求出这个值. 先说复杂度最高的:O(n3) 直接上代码,非常easy的: // // ...
- 动态规划 Common Subsequence
描述 A subsequence of a given sequence is the given sequence with some elements (possible none) left o ...
- Common Subsequence 最大公共子序列问题
Problem Description A subsequence of a given sequence is the given sequence with some elements (poss ...
随机推荐
- rabbitmq 交换机模式 -主题模式 topic
建立一个交换机 tpc 并且绑定了各自的路由到 Q1 Q2 <?php require_once "./vendor/autoload.php"; use PhpAmqpLi ...
- ansible用get_url模块在受控机下载文件(ansible2.9.5)
一,ansible的get_url模块用途: get_url模块可以在受控机下载文件 可以理解成从受控端执行wget 下载的url支持:http | https | ftp 三种协议 说明:刘宏缔 ...
- laravel重写
laravel location / { try_files $uri $uri/ /index.php?$query_string; } ci location / { try_files $uri ...
- day71:drf:API接口&Restful API规范&Django Rest Framework&drf中的序列化和反序列化功能
目录 1.web应用模式 2.API接口 3.Restful API规范 4.序列化 5.Django Rest Framework 1.drf的简单介绍 2.drf的特点 3.如何安装drf 4.d ...
- linux上安装mitmproxy
一.去git上下载安装包 下载mitmproxy二进制安装包:https://github.com/mitmproxy/mitmproxy/releases/ 二.安装 #上传 rz 安装包的本地路径 ...
- Vue.js 3.0搭配.NET Core写一个牛B的文件上传组件
在开发Web应用程序中,文件上传是经常用到的一个功能. 在Jquery时代,做上传功能,一般找jQuery插件就够了,很少有人去探究上传文件插件到底是怎么做的. 简单列一下我们要做的技术点和功能点 使 ...
- Kitty基于Spring Boot、Spring Cloud、Vue.js、Element实现前后端分离的权限管理系统
源码地址:https://gitee.com/liuge1988/kitty 软件架构 后端架构 开发环境 IDE : eclipse 4.x JDK : JDK1.8.x Maven : Maven ...
- 前后端分离Java后端主流开发环境框架20200622
开发环境: IDE:IntelliJ IDEA 2017+ DB: mysql5.7.4.PostgreSQL.mongoDB.redis JDK:JDK1.8+ Maven:Maven 3.2.3+ ...
- 《Head First 设计模式》:与设计模式相处
正文 一.设计原则 1.封装变化 找出应用中可能需要变化之处,把它们独立出来,不要和那些不需要变化的代码混在一起. 2.针对接口编程,不针对实现编程 "针对接口编程"真正的意思是& ...
- Docker(1)- 什么是 Docker
如果你还想从头学起 Docker,可以看看这个系列的文章哦! https://www.cnblogs.com/poloyy/category/1690628.html 备注 这里的概念直接引用官方的, ...