最长公共子串 Longest Common Subsequence

动态规划的经典题！

package DP;

import java.util.Arrays;

// 最长公共子串 Longest Common Subsequence
public class LCS {

	static int dp[][] = null;
	
	public static void main(String[] args) {
		String a = "ABCABCBAASJKDFHSDDSJSAHJSD";
		String b = "CBABCABCCSDESJASDFHSDFSW";
		
		dp = new int[a.length()+1][b.length()+1];
		for(int[] row : dp){
			Arrays.fill(row, -1);
		}
		
		System.out.println(lcs2(a.toCharArray(), a.length(), b.toCharArray(), b.length()));
		System.out.println(lcs3(a.toCharArray(), a.length(), b.toCharArray(), b.length()));
//		print();
		System.out.println(lcs(a.toCharArray(), a.length(), b.toCharArray(), b.length()));
		
	}

	// 纯递归O(m*2^n)
	public static int lcs(char[] A, int m, char[] B, int n){
		if(m==0 || n==0){
			return 0;
		}
		
		if(A[m-1] == B[n-1]){
			return 1 + lcs(A, m-1, B, n-1);
		}else{
			return Math.max(lcs(A, m, B, n-1), lcs(A, m-1, B, n));
		}
	}
	
	// DP, top-down O(n^2)
	public static int lcs2(char[] A, int m, char[] B, int n){
		if(m==0 || n==0){
			return 0;
		}
		
		// 如果已经存在dp数组中，直接返回
		if(dp[m][n] != -1){
			return dp[m][n];
		}
		int res = 0;
		if(A[m-1] == B[n-1]){
			res = 1 + lcs2(A, m-1, B, n-1);
		}else{
			res = Math.max(lcs2(A, m, B, n-1), lcs2(A, m-1, B, n));
		}
		dp[m][n] = res;		// 把新值记录到dp数组中
		return res;
	}
	
	// DP, bottom-up O(n^2)
	public static int lcs3(char[] A, int m, char[] B, int n){
		for(int i=0; i<=m; i++){
			for(int j=0; j<=n; j++){
				if(i==0 || j==0){
					dp[i][j] = 0;
				}
				else if(A[i-1] == B[j-1]){
					dp[i][j] = dp[i-1][j-1] + 1;
				}
				else{
					dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
				}
			}
		}
		return dp[m][n];
	}
	
	public static void print(){
		for(int i=0; i<dp.length; i++){
			for(int j=0; j<dp[0].length; j++){
				System.out.print(dp[i][j] + " ");
			}
			System.out.println();
		}
	}
}

Ref:

http://www.csie.ntnu.edu.tw/~u91029/LongestCommonSubsequence.html

http://www.geeksforgeeks.org/dynamic-programming-set-4-longest-common-subsequence/

http://techieme.in/techieme/dynamic-programming-longest-common-subsequence/

http://www.cc.gatech.edu/~ninamf/Algos11/lectures/lect0311.pdf

http://blog.gaurav.im/?p=9

http://blog.csdn.net/alexander_xfl/article/details/11556655