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原题链接在这里：http://www.lintcode.com/en/problem/longest-common-substring/#
题目：
Given two strings, find the longest common substring.
Return the length of it.
Given A = "ABCD", B = "CBCE", return 2. "BC"
O(n x m) time and memory.
题解：
与Longest Common Subsequence相似.
DP. 参考http://www.geeksforgeeks.org/longest-common-substring/
dp[i][j]表示长度为i的str1 和 长度为j的str2 longest common substring长度.
看两个substring 的 suffix.
dp[i][j] = dp[i-1][j-1] + 1 if str1.charAt(i-1) == str2.charAt(j-1).
dp[i][j] = 0. o.w
Time Complexity: O(m*n). Space: O(m*n).
AC Java:
1 public class Solution { 2 /** 3 * @param A, B: Two string. 4 * @return: the length of the longest common substring. 5 */ 6 public int longestCommonSubstring(String A, String B) { 7 if(A == null || B == null){ 8 return 0; 9 } 10 int m = A.length(); 11 int n = B.length(); 12 int [][] dp = new int[m+1][n+1]; 13 int longest = 0; 14 for(int i = 1; i<=m; i++){ 15 for(int j = 1; j<=n; j++){ 16 if(A.charAt(i-1) == B.charAt(j-1)){ 17 dp[i][j] = dp[i-1][j-1]+1; 18 }else{ 19 dp[i][j] = 0; 20 } 21 longest = Math.max(longest, dp[i][j]); 22 } 23 } 24 return longest; 25 } 26 }
LintCode Longest Common Substring
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5349760.html
