22
33/**
44 * 动态规划 dp[i + 1][j + 1] = Math.max(dp[i+1][j], dp[i][j+1]) o(m*n)
5- *
5+ * <p>
66 * 若题目为最长公共子串,则在c1,c2不相等时不做处理(赋值0),在遍历过程中记录最大值即可
77 */
88public class Solution {
@@ -25,4 +25,37 @@ public int longestCommonSubsequence(String text1, String text2) {
2525 }
2626 return dp [m ][n ];
2727 }
28+ 29+ /**
30+ * 最长公共字串
31+ *
32+ * @param str1
33+ * @param str2
34+ * @return
35+ */
36+ public static String longestCommonSubstring (String str1 , String str2 ) {
37+ int m = str1 .length ();
38+ int n = str2 .length ();
39+ int [][] dp = new int [m + 1 ][n + 1 ];
40+ int maxLength = 0 ;
41+ int endIndex = -1 ;
42+ 43+ for (int i = 1 ; i <= m ; i ++) {
44+ for (int j = 1 ; j <= n ; j ++) {
45+ if (str1 .charAt (i - 1 ) == str2 .charAt (j - 1 )) {
46+ dp [i ][j ] = dp [i - 1 ][j - 1 ] + 1 ;
47+ if (dp [i ][j ] > maxLength ) {
48+ maxLength = dp [i ][j ];
49+ endIndex = i - 1 ;
50+ }
51+ } else {
52+ dp [i ][j ] = 0 ;
53+ }
54+ }
55+ }
56+ if (maxLength == 0 ) {
57+ return "" ;
58+ }
59+ return str1 .substring (endIndex - maxLength + 1 , endIndex + 1 );
60+ }
2861}
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