动态规划：最长公共子串长度

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设M=str1.length(),N=str2.length();

时间复杂度O(M*N),额外空间复杂度也是O(M*N).

建立一个M*N矩阵dp。 其中dp[i][j]表示在必须把str[i]和str2[j]当做公共子串最后一个字符的情况下，公共子串最长能多长。

JAVA代码：

<span style="font-size:18px;">import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();

		while (N-- > 0) {
			String str1 = sc.next();
			String str2=sc.next();
			//StringBuffer sb = new StringBuffer(str1);
			//String str2 = sb.reverse().toString();
			int length1 = longSubString(str1, str2);
			//System.out.println(str1.length() - length1);
			 System.out.println(length1);
		}
	}

	public static int longSubString(String str1, String str2) {
		int max = 0;
		int m = str1.length();
		int n = str2.length();
		int dp[][] = new int[m][n];
		for (int i = 0; i < m; i++)
			if (str1.charAt(i) == str2.charAt(0)) {
				dp[i][0] = 1;
				max = 1;
			}
		for (int i = 0; i < n; i++)
			if (str2.charAt(i) == str1.charAt(0)) {
				dp[0][i] = 1;
				max = 1;
			}

		for (int i = 1; i < m; i++) {
			for (int j = 1; j < n; j++) {
				if (str1.charAt(i) == str2.charAt(j)) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
					max = Math.max(max, dp[i][j]);
				}
			}
		}
		return max;
	}</span>