编程之美--3.3(2)最长公共子串

求两个字符串的最长公共子串。例如 strA="dabcabd" strB = "dbabcaeefbd" 最长公共子串为"abca"

方法：类似于编程之美3.3http://www.cnblogs.com/gnivor/articles/4604733.html

采用动态规划的方法。设定一个数组c[lenA+1][lenB+1],第0行和第0列均赋值为0。

动态转移方程为：
【1】如果strA.charAt(j - 1) == strB.charAt(i - 1)，那么c[i][j]=c[i-1][j-1]+1
【2】如果strA.charAt(j - 1) != strB.charAt(i - 1)，那么c[i][j]=0;
注意：c[i][j]第0行第0列不记录数组内的信息，且第0行和第0列实质上只用到了c[0][0]
然后在数组中找到最大的那个数，沿着对角线方向，向前一直推即可得到最长公共子串

public static String longestSameSubstring(String strA ,String strB) {
    int lenA = (int) strA.length();
    int lenB = (int) strB.length();
    int maxi = 0;
    int maxlen = 0;
    //c[i][j]存放的为子串相似的数量。（因为c的边界不算）
    int[][] c = new int[lenA + 1][lenB + 1];
    // Record the distance of all begin points of each String
    // 初始化边界
    for (int i = 0; i <= lenA; i++)
        c[i][0] = 0;
    for (int j = 0; j <= lenB; j++)
        c[0][j] = 0;
    c[0][0] = 0;
    //递推求出所有c元素
    for (int i = 1; i <= lenA; i++)
        for (int j = 1; j <= lenB; j++) {
            if (strB.charAt(j - 1) == strA.charAt(i - 1)){
                c[i][j] = c[i - 1][j - 1]+1;
                if(c[i][j]>maxlen){
                    maxi = i;
                    maxlen=c[i][j];
                }
            }
            else
                c[i][j] = 0;
        }
    for (int i = 0; i <= lenA; i++) {
        for (int j = 0; j <= lenB; j++) {
            System.out.print(c[i][j] + " ");
        }
        System.out.println();
    }
    System.out.println("最大长度"+maxlen);
    int start = maxi-maxlen;
    return (String) strA.subSequence(start, maxi);    
}

main()

String str = longestSameSubstring("dabcabd","dbabcaeefbd");
System.out.println("最长公共子串为"+str);

结果：

0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 1 0 0 0 0 0
0 0 1 0 2 0 0 0 0 0 1 0
0 0 0 0 0 3 0 0 0 0 0 0
0 0 0 1 0 0 4 0 0 0 0 0
0 0 1 0 2 0 0 0 0 0 1 0
0 1 0 0 0 0 0 0 0 0 0 2
最大长度4
最长公共子串为abca