F面经prepare:strstr变种


F面经prepare:strstr变种
 * Given an integer k>=1 and two strings A and B (length ~n each);
 * Figure out if there is any common substring of length at least k.
 * (i.e. any string of length at least k, that is a substring of both A and B)
 *
 *  A="facebook", B="bookshelf", k=3  ==> true
           ^^^       ^^^
 *  A="facebook", B="bookshelf", k=4  ==> true
           ^^^^      ^^^^
 *  A="facebook", B="bookshelf", k=5  ==> false
 1 public boolean check(String A, String B, int k) {
 2      
 3      int lenA = A.length();
 4      int lenB = B.length();
 5      for (int i=0; i<=lenA-k; i++) {
 6          String stra = A.substring(i, i+k);
 7          for (int j=0; j<=lenB-k; j++) {
 8              String strb = B.substring(j,j+k);
 9              if (stra.equals(strb)) return true;
10          }
11      }
12      return false;
13  }
follow up: How to optimize?
想到KMP了,trie了,居然没有想到HashSet,然后KMP和trie的时间复杂度又没有搞太对
他们真的很喜欢问时间复杂度,空间复杂度,时间换空间,空间换时间
 1 public boolean check(String A, String B, int k) {
 2     //store B's substring of length k to hashSet
 3     HashSet<String> set = new HashSet<String>();
 4     for (int i=0; i<B.length()-k; i++) {
 5         set.add(B.substring(i, i+k));
 6     }           
 7     for (int i=0; i<A.length()-k; i++) {
 8         String sbstra = A.substring(i, i+k); // don't use substring
 9         if (B.contains(sbstra)) return true;
10     }
11     return false;
12 }
这道题一点都不难,只是
1.受以前思维定式影响
2. 考场真的是思路就很局限, 放不开
3. 这道题面经还见过,别人提到过set,没细看,恰好这个方法忘记了
4. 考场写的时候,真的是连for(int i=0; i<=A.length()-k; i++) 这个是不是A.length()-k都想不清楚

应对:
1.重视面经。看过做过会占很大便宜
2. 要加强时间复杂度,空间复杂度训练
3. 要加强思维敏捷度训练
4. 写code能力(比如这次A.length()-k)
F面经prepare:strstr变种

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