题解 | #公共子串计算#
最长公共子串，同 https://www.nowcoder.com/practice/181a1a71c7574266ad07f9739f791506 类似，只是此题只输出长度（上一题输出子串）
可以用动态规划，可以用双指针。
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str1 = sc.nextLine();
String str2 = sc.nextLine();
String minStr = str1.length() <= str2.length() ? str1 : str2;
String maxStr = str1.length() > str2.length() ? str1 : str2;
helper1(minStr, maxStr);
helper2(minStr, maxStr);
}
/**
* 最长公共子串问题，利用动态规划解决
* dp[i][j]表示A以第i个字符结尾，B以第j个字符结尾，最长公共子串长度
* 若A[i] == B[j]，则dp[i][j] = dp[i-1][j-1];
* 若A[i] != B[j]，则dp[i][j] = 0;
*/
public static void helper1(String minStr, String maxStr) {
int[][] dp = new int[minStr.length()+1][maxStr.length()+1];
int maxLen = 0;
for (int i = 1; i <= minStr.length(); i++) {
for (int j = 1; j <= maxStr.length(); j++) {
if (minStr.charAt(i) == maxStr.charAt(j)) {
dp[i][j] = dp[i-1][j-1] + 1;
maxLen = Math.max(maxLen, dp[i][j]);
} else {
dp[i][j] = 0;
}
}
}
System.out.println(maxLen);
}
public static void helper2(String minStr, String maxStr) {
int result = 0;
int left = 0, right = left + 1;
while (right <= minStr.length()) {
String tmp = minStr.substring(left, right);
if (maxStr.contains(tmp)) {
result = Math.max(result, tmp.length());
right++;
} else {
left++;
right = left + 1;
}
}
System.out.println(result);
}
}
