题解 | #最长公共子串#
最长公共子串
https://www.nowcoder.com/practice/f33f5adc55f444baa0e0ca87ad8a6aac
import java.util.*;
public class Solution {
public String LCS (String str1, String str2) {
int m = str1.length(), n = str2.length();
String[][] dp = new String[m + 1][n + 1];
String max = "";
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if(i == 0 || j == 0){
dp[i][j] = "";
}else if (str1.charAt(i-1) == str2.charAt(j-1)) {
dp[i][j] = dp[i-1][j-1] + str1.charAt(i-1);
if (max.length() < dp[i][j].length()) {
max = dp[i][j];
}
}else{
dp[i][j] = ""; //与最长公共子序列不同 这里不连续的话直接为空，
}
}
}
return max.length() == 0 ? "-1" : max;
}
}
代码思路与最长公共子序列一致，只有一行代码不同，这样比较好记忆
