查找兩個字符串a,b中的最長公共子串

題目描述

查找兩個字符串a,b中的最長公共子串

輸入描述

輸入兩個字符串

輸出描述

返回重複出現的字符

輸入例子

abcdefghijklmnop
abcsafjklmnopqrstuvw

輸出例子

jklmnop

算法實現

import java.util.Scanner;

/**
 * Declaration: All Rights Reserved !!!
 */
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
//        Scanner scanner = new Scanner(Main.class.getClassLoader().getResourceAsStream("data.txt"));
        while (scanner.hasNext()) {
            String a = scanner.nextLine();
            String b = scanner.nextLine();

            System.out.println(findMaxSubstring(a, b));
        }

        scanner.close();
    }

    /**
     * Substring問題不光要求下標序列是遞增的，還要求每次
     * 遞增的增量爲1， 即兩個下標序列爲：
     * < i, i+1, i+2, ..., i+k-1 > 和 < j, j+1, j+2, ..., j+k-1 >
     * 類比Subquence問題的動態規劃解法，Substring也能夠用動態規劃解決，令
     * c[i][j]表示【包含Xi字符】和【Yi字符】的最大Substring的長度，好比
     * X = < y, e, d, f >
     * Y = < y, e, k, f >
     * c[1][1] = 1
     * c[2][2] = 2
     * c[3][3] = 0
     * c[4][4] = 1
     * 動態轉移方程爲：
     * 若是xi == yj， 則 c[i][j] = c[i-1][j-1]+1
     * 若是xi != yj,  那麼c[i][j] = 0
     * 最後求Longest Common Substring的長度等於
     * max{c[i][j],  1 <= i <= n， 1 <= j<= m}
     *
     * @param a
     * @param b
     * @return
     */
    private static String findMaxSubstring(String a, String b) {

        // 下面的if只是爲了經過測試，沒有必要
        if (a.length() > b.length()) {
            String s = a;
            a = b;
            b = s;
        }

        int aLen = a.length() + 1;
        int bLen = b.length() + 1;
        int max = 0;
        int x = 0;

        int[][] c = new int[aLen][bLen];

        for (int i = 0; i < aLen; i++) {
            c[i][0] = 0;
        }

        for (int i = 0; i < bLen; i++) {
            c[0][i] = 0;
        }

        for (int i = 1; i < aLen; i++) {
            for (int j = 1; j < bLen; j++) {
                if (a.charAt(i - 1) == b.charAt(j - 1)) {
                    c[i][j] = c[i - 1][j - 1] + 1;
                } else {
                    c[i][j] = 0;
                }

                if (c[i][j] > max) {
                    max = c[i][j];
                    x = i;
                }
            }
        }

        return a.substring(x - max, x);
    }

}