题解 | 最长公共子串

import java.util.*;
import static java.lang.Math.max;

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * longest common substring
     * @param str1 string字符串 the string
     * @param str2 string字符串 the string
     * @return string字符串
     */
    public String LCS (String str1, String str2) {
        // write code here

        int x = str1.length();
        int y = str2.length();
        if (x == 0 || y == 0) {
            //return "";
            return "";
        }
        int[][] arr = new int[x + 1][y + 1];
        int[][] arrnew = new int[x + 1][y + 1];
        for (int i = 0; i <= x; i++) {
            for (int j = 0; j <= y; j++) {
                arr[i][j] = 0;
                arrnew[i][j] = 0;
            }
        }

        for (int i = 1; i <= x; i++) {
            for (int j = 1; j <= y; j++) {
                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    arr[i][j] = 1;
                } else {
                    arr[i][j] = 0;
                }
            }
        }
        //由arr-》arrnew
	  	//判断当前字符节点相等并且上一个节点也需要相等，才能➕1，否则只能为1
        for (int i = 1; i <= x; i++) {
            for (int j = 1; j <= y; j++) {
                if (arr[i][j] == 1 && arr[i - 1][j - 1] == 1) {
                    arrnew[i][j] = max(max(arrnew[i - 1][j - 1] + 1, arrnew[i - 1][j]),
                                       arrnew[i][j - 1]);
                }
                if (arr[i][j] == 1 && arr[i - 1][j - 1] == 0) {
                    arrnew[i][j] = max(max( 1, arrnew[i - 1][j]), arrnew[i][j - 1]);
                }

            }
        }


        int max1 = 0;

        for (int i = 1; i <= x; i++) {
            for (int j = 1; j <= y; j++) {
                max1 = max(max1, arrnew[i][j]);
            }
        }
    


        int res = -1;
        String muban1 = "";
        for(int i=0;i<=x-max1;i++) {
            muban1 = str1.substring(i, i + max1);
            for (int j = 0; j <= y - max1; j++) {
                String muban2 = str2.substring(j, j + max1);
                if (muban1.equals(muban2)) {
                    //return muban1;
                    res=1;
                    break;
                }
            }
            if(res == 1){
                break;
            }
        }
     
        return muban1;

    }

}