Leetcode: 132 Pattern
Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list. Note: n will be less than 15,000. Example 1: Input: [1, 2, 3, 4] Output: False Explanation: There is no 132 pattern in the sequence. Example 2: Input: [3, 1, 4, 2] Output: True Explanation: There is a 132 pattern in the sequence: [1, 4, 2]. Example 3: Input: [-1, 3, 2, 0] Output: True Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
我觉得这道题是hard,难点第一是要想到用stack,第二是要维护一个这样子的min-max序列:So at any time in the stack, non-overlapping Pairs
are formed in descending order by their min value, which means the min value of peek element in the stack is always the min value globally.
The idea is that we can use a stack to keep track of previous min-max intervals.
For each number num
in the array
If stack is empty:
- push a new Pair of
num
into stack
If stack is not empty:
-
if
num
<stack.peek().min
, push a new Pair ofnum
into stack -
if
num
>=stack.peek().min
, we first pop() out the peek element, denoted aslast
-
if
num
<last.max
, we are done, returntrue
; -
if
num
>=last.max
, we mergenum
intolast
, which meanslast.max
=num
.
Once we updatelast
, if stack is empty, we just push backlast
.
However, the crucial part is:
If stack is not empty, the updatedlast
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 are done ,return
true
- Entirely covered stack.peek(), i.e.
-
refer to: https://discuss.leetcode.com/topic/68193/java-o-n-solution-using-stack-in-detail-explanation/2
我在它基础上稍作改动
1 public class Solution { 2 public class Pair { 3 int min; 4 int max; 5 public Pair(int n1, int n2) { 6 min = n1; 7 max = n2; 8 } 9 } 10 11 public boolean find132pattern(int[] nums) { 12 if (nums.length < 3) return false; 13 Stack<Pair> st = new Stack<Pair>(); 14 for (int n : nums) { 15 if (st.isEmpty() || n<=st.peek().min) st.push(new Pair(n, n)); 16 else { 17 if (n < st.peek().max) return true; 18 Pair last = st.pop(); 19 last.max = Math.max(last.max, n); 20 while (!st.isEmpty() && last.max>=st.peek().max) st.pop(); 21 if (!st.isEmpty() && last.max>st.peek().min) return true; 22 st.push(last); 23 } 24 } 25 return false; 26 } 27 }