LintCode: Longest Common Substring

C++

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两个游标一个长度

i遍历a

j遍历b

len遍历公共子串长度

 1 class Solution {
 2 public:    
 3     /**
 4      * @param A, B: Two string.
 5      * @return: the length of the longest common substring.
 6      */
 7     int longestCommonSubstring(string &A, string &B) {
 8         // write your code here
 9         int m = A.size();
10         int n = B.size();
11         if (m == 0 || n == 0) {
12             return 0;
13         }
14         int ans = 0;
15         for (int i = 0; i < m; ++i) {
16             for (int j = 0; j < n; ++j) {
17                 int len = 0;
18                 while (i + len < m && j + len < n && A[i + len] == B[j + len]) {
19                     len ++;
20                     ans = max(ans, len);
21                 }
22             }
23         }
24         return ans;
25     }
26 };

C++,

dp

 1 class Solution {
 2 public:    
 3     /**
 4      * @param A, B: Two string.
 5      * @return: the length of the longest common substring.
 6      */
 7     int longestCommonSubstring(string &A, string &B) {
 8         // write your code here
 9         int m = A.size();
10         int n = B.size();
11         if (m == 0 || n == 0) {
12             return 0;
13         }
14         vector<vector <int> > dp(m + 1, vector<int>(n + 1));
15         // int[][] dp = new int[m + 1][n + 1];
16         int ans = 0;
17         for (int i = 0; i <= m; ++i) {
18             for (int j = 0; j <= n; ++j) {
19                 if (i == 0 || j == 0) {
20                     dp[i][j] = 0;
21                 } else {
22                     if (A[i-1] == B[j-1]) {
23                         dp[i][j] = dp[i - 1][j - 1] + 1;
24                         ans = max(dp[i][j], ans);
25                     } else {
26                         dp[i][j] = 0;
27                     }
28                 }
29             }
30         }
31         return ans;
32     }
33 };

 

posted @ 2015-12-01 12:18  ZH奶酪  阅读(301)  评论(0)  编辑  收藏  举报