输入：s1 = "photograph" s2 = "tomography"
输出：ograph
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二、解题思路

​ 其实本题就是字符串编辑距离的变种，字符串编辑距离会了，此题就会了。本题跟最长公共子序列也是非常像。详见动态规划经典题目-最长公共子序列

1. 定义状态

​ 设dp[i][j]表示字符串s1的前i个字符与字符串s2前j个字符的最长公共子串的长度。

2. 定义状态转移方程

当s1[i] == s2[j]时，有$$dp[i][j]=dp[i-1][j-1]+1$$

否则$$dp[i][j]=0$$

3. 初始化

​ 显然dp[0][j]初始化为0。

​ 显然dp[i][0]初始化为0。

三、代码实现

/**
 * 求两个字符串的最长公公子串
 *
 *
 *  @author hh
 *  @date 2021-5-16 23:08
 */
public class LongestCommonSubString {

    public String longestCommonSubString(String s1,String s2){
        int[][] dp = new int[s1.length() + 1][s2.length() + 1];
        //初始化行
        for(int i = 0 ; i <= s2.length(); i++){
            dp[0][i] = 0;
        }
        //初始化列
        for(int i = 0; i <= s1.length(); i++){
            dp[i][0] = 0;
        }
        //计算dp数组
        for(int i = 1; i <= s1.length(); i++){
            for(int j = 1; j <= s2.length(); j++){
                if(s1.charAt(i - 1) == s2.charAt(j -1 )){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }else{
                    dp[i][j] = 0;
                }
            }
        }
        //匹配的最长子串长度
        int maxLength = Integer.MIN_VALUE;
        int index = 0;
        //检测是否匹配
        for(int i = 0; i <= s2.length(); i++){
            if(dp[s1.length()][i] > maxLength){
                maxLength = dp[s1.length()][i];
                index = i;
            }
        }
        return s2.substring(index - maxLength,index);
    }

    public static void main(String[] args){
        LongestCommonSubString longestCommonSubString = new LongestCommonSubString();
        String s1 = "photograph";
        String s2 = "tomography";
        System.out.println(longestCommonSubString.longestCommonSubString(s1,s2));
    }
}

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