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); } }