本文介绍了两个字符串的所有可能的LCS(最长公共子)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我们可以发现LCS两个字符串(最长公共子)与DP(动态规划)。通过跟踪与DP表我们可以得到的LCS。但是,如果存在多个LCS怎样才能让所有的人?
We can find the LCS(Longest Common Subsequence) of two strings with DP(Dynamic Programming). By keeping track with DP Table we can get the LCS. But if there exists more than one LCS how can we get all of them?
例如:
string1 : bcab
string2 : abc
这里既有AB和BC是濒海战斗舰。
Here both "ab" and "bc" are LCS.
推荐答案
下面是一个Java的解决方案。为了说明你可以看到我的回答How打印的最长公共子序列的所有可能的解决方案
Here is a working java solution. For explanation you can see my answerHow to print all possible solutions for Longest Common subsequence
static int arr[][];
static void lcs(String s1, String s2) {
for (int i = 1; i <= s1.length(); i++) {
for (int j = 1; j <= s2.length(); j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1))
arr[i][j] = arr[i - 1][j - 1] + 1;
else
arr[i][j] = Math.max(arr[i - 1][j], arr[i][j - 1]);
}
}
}
static Set<String> lcs(String s1, String s2, int len1, int len2) {
if (len1 == 0 || len2 == 0) {
Set<String> set = new HashSet<String>();
set.add("");
return set;
}
if (s1.charAt(len1 - 1) == s2.charAt(len2 - 1)) {
Set<String> set = lcs(s1, s2, len1 - 1, len2 - 1);
Set<String> set1 = new HashSet<>();
for (String temp : set) {
temp = temp + s1.charAt(len1 - 1);
set1.add(temp);
}
return set1;
} else {
Set<String> set = new HashSet<>();
Set<String> set1 = new HashSet<>();
if (arr[len1 - 1][len2] >= arr[len1][len2 - 1]) {
set = lcs(s1, s2, len1 - 1, len2);
}
if (arr[len1][len2 - 1] >= arr[len1 - 1][len2]) {
set1 = lcs(s1, s2, len1, len2 - 1);
}
for (String temp : set) {
set1.add(temp);
}
//System.out.println("In lcs" + set1);
return set1;
}
}
public static void main(String[] args) {
String s1 = "bcab";
String s2 = "abc";
arr = new int[s1.length() + 1][s2.length() + 1];
lcs(s1, s2);
System.out.println(lcs(s1, s2, s1.length(), s2.length()));
}
这篇关于两个字符串的所有可能的LCS(最长公共子)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!