leetcode:132 Pattern

 Given a sequence of n integers a1, a2, ..., an, a-pattern is a subsequence ai, AJ, AK such that I < J < K A nd AI < ak < AJ.  Design an algorithm, takes a list of n numbers as input and checks whether there is a 132 The pattern in the list. Note:n'll is less than  15,000 1:input: [ 1, 2, 3, 4]output: Falseexplanation:there is no  132 pattern in the sequence. Example  2:input: [ 3, 1, 4, 2]output: Trueexplanation:there is a  pattern in the sequence: [1, 4, 2 3:input: [-1, 3, 2, 0]output: Trueexplanation:there is three  patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0]. 

I think this problem is hard, the difficulty first is to think of stack, the second is to maintain a min-max sequence such as: So at any time in the stack, non-overlapping is Pairs formed I n Descending order by their min value, which means the min value of Peek element in the stack was always the min value Glob Ally.

The idea is, we can use a stack to keep track of previous min-max intervals.

For each number in the num array

If Stack is empty:

• Push a new Pair of into num stack

If stack is not empty:

• If num < stack.peek().min , push a new Pair of into num stack

• If num >= stack.peek().min , we first pop () out the peek element, denoted aslast

  • if  num  <  Last.max , we is done, return  true ;

  • if  num  >=  Last.max , we merge  num  into   last , which means  Last.max  =  num .
    Once we update  last , if stack was empty, we just push back  last .
    However, the crucial part was:
    If Stack is isn't empty, the updated  last   Might:

    • entirely covered stack.peek (), i.e.  last.min  <  Stack.peek ( ). Min   (which is always true) &&  Last.max  >=  Stack.peek (). Max , in which case we keep popping out Stack.peek ().
    • Form A 1-3-2 pattern, we is done, return  true

Refer TO:HTTPS://DISCUSS.LEETCODE.COM/TOPIC/68193/JAVA-O-N-SOLUTION-USING-STACK-IN-DETAIL-EXPLANATION/2

1 classpair{2         intmin, Max;3          PublicPair (intMinintmax) {4              This. Min =min;5              This. Max =Max;6         }7     }8      Public BooleanFind132pattern (int[] nums) {9stack<pair> stack =NewStack ();Ten          for(intn:nums) { One             if(Stack.isempty () | | n <stack.peek (). Min) Stack.push (NewPair (n,n)); A             Else if(N >Stack.peek (). Min) {  -Pair last =Stack.pop (); -                 if(N < Last.max)return true; the                 Else { -Last.max =N; -                      while(!stack.isempty () && n >=Stack.peek (). Max) Stack.pop (); -                     //at this time, n < Stack.peek (). Max (if stack is not empty) +                     if(!stack.isempty () && Stack.peek (). Min < N)return true; - Stack.push (last); +                 } A                  at             } -         } -         return false; -}

My method, line 15th, is different.

1  Public classSolution {2      Public classPair {3         intmin;4         intMax;5          PublicPair (intN1,intn2) {6Min =N1;7Max =N2;8         }9     }Ten      One      Public BooleanFind132pattern (int[] nums) { A         if(Nums.length < 3)return false; -Stack<pair> st =NewStack<pair>(); -          for(intn:nums) { the             if(St.isempty () | | n<=st.peek (). Min) St.push (NewPair (n, N)); -             Else { -                 if(N < St.peek (). Max)return true; -Pair last =St.pop (); +Last.max =Math.max (Last.max, n); -                  while(!st.isempty () && last.max>=St.peek (). Max) St.pop (); +                 if(!st.isempty () && Last.max>st.peek (). Min)return true; A St.push (last); at             } -         } -         return false; -     } -}

leetcode:132 Pattern