Interview-Increasing Sequence with Length 3.

LiBlog發表於2014-12-24
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Given an array, determine whether there are three elements A[i],A[j],A[k], such that A[i]<A[j]<A[k] & i<j<k.

Analysis:

It is a special case of the Longest Increasing Sequence problem. We can use the O(nlog(n)) algorithm, since we only need sequence with length three, we can do it in O(n).

Solution:

 1 public static boolean threeIncSeq(int[] A){
 2         if (A.length<3) return false;
 3 
 4         int oneLen = 0;
 5         int twoLen = -1;
 6         for (int i=1;i<A.length;i++){
 7             //check whether current element is larger then A[twoLen].
 8             if (twoLen!=-1 && A[i]>A[twoLen]) return true;
 9             if (twoLen!=-1 && A[i]>A[twoLen] && A[i]<A[oneLen]){
10                 twoLen = i;
11                 continue;
12             }
13             if (twoLen==-1 && A[i]>A[oneLen]){
14                 twoLen = i;
15                 continue;
16             }
17             if (A[i]<A[oneLen]){
18                 oneLen = i;
19                 continue;
20             }
21         }
22 
23         return false;
24     }

Variant:

If we need to output the sequence, we then need to record the sequence of current length 1 seq and length 2 seq.

 1 public static List<Integer> threeIncSeq(int[] A){
 2         if (A.length<3) return (new ArrayList<Integer>());
 3 
 4         int oneLen = 0;
 5         int twoLen = -1;
 6         List<Integer> oneList = new ArrayList<Integer>();
 7         List<Integer> twoList = new ArrayList<Integer>();
 8         oneList.add(A[0]);
 9         for (int i=1;i<A.length;i++){
10             //check whether current element is larger then A[twoLen].
11             if (twoLen!=-1 && A[i]>A[twoLen]){
12                 twoList.add(A[i]);
13                 return twoList;
14             }
15             if (twoLen!=-1 && A[i]>A[twoLen] && A[i]<A[oneLen]){
16                 twoLen = i;
17                 twoList = new ArrayList<Integer>();
18                 twoList.addAll(oneList);
19                 twoList.add(A[i]);
20                 continue;
21             }
22             if (twoLen==-1 && A[i]>A[oneLen]){
23                 twoLen = i;
24                 twoList = new ArrayList<Integer>();
25                 twoList.addAll(oneList);
26                 twoList.add(A[i]);
27                 continue;
28             }
29             if (A[i]<A[oneLen]){
30                 oneLen = i;
31                 oneList = new ArrayList<Integer>();
32                 oneList.add(A[i]);
33                 continue;
34             }
35         }
36 
37         return (new ArrayList<Integer>());
38 
39     }

NOTE: This is more compliated then needed, when using List<> in this case, but this idea can be used to print the LIS.

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