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].
一刷
題解：
利用棧來(lái)保存之前的min-max intervals
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 of num into stack
-
if num >= stack.peek().min, we first pop() out the peek element, denoted as last
if num < last.max, we are done, return true;
-
if num >= last.max, we merge num into last, which means last.max = num.
Once we update last, if stack is empty, we just push back last.
However, the crucial part is:
If stack is not 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 are done ,return true
目的：保證stack里面保存的pair是無(wú)overlap的，且是按照min-value遞減的。所以peek()處的min為全局最小值屡贺。
public class Solution {
class Pair{
int min, max;
public Pair(int min, int max){
this.min = min;
this.max = max;
}
}
public boolean find132pattern(int[] nums) {
Stack<Pair> stack = new Stack<>();
for(int num : nums){
if(stack.isEmpty() || num<stack.peek().min) stack.push(new Pair(num, num));
else if (num>stack.peek().min){
Pair last = stack.pop();
if(num < last.max) return true;
last.max = num;
while(!stack.isEmpty() && num >= stack.peek().max) stack.pop();
if(!stack.isEmpty() && stack.peek().min<num) return true;
stack.push(last);
}
}
return false;
}
}
