Given a sequence of integers: a1, a2, …, an, a subsequence ai, aj, ak of 132 mode is defined as: when i <j <k, ai <ak <aj. Design an algorithm, when given a sequence of n numbers, verify whether the sequence contains a sub-sequence of 132 patterns.
Note: The value of n is less than 15000.
Example 1:
Input: [1, 2, 3, 4]
Output: False
Explanation: There is no subsequence of 132 mode in the sequence.
Example 2:
Input: [3, 1, 4, 2]
Output: True
Explanation: There is a subsequence of 132 pattern in the sequence: [1, 4, 2].
Example 3:
Input: [-1, 3, 2, 0]
Output: True
Explanation: There are 3 subsequences of 132 pattern in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
代码：
public boolean find132pattern(int[] nums) {
if(nums.length<3) {
return false;
}
for(int i=1;i<nums.length-1;i++) {
int min=Integer.MAX_VALUE;
for(int j=0;j<i;j++) {
if(nums[j]<nums[i]&&min>nums[j]) {
min=nums[j];
}
}
if(min==Integer.MAX_VALUE) {
continue;
}
for(int j=i+1;j<nums.length;j++) {
if(nums[j]<nums[i]&&nums[j]>min) {
return true;
}
}
}
return false;
}
学习大佬代码：
public boolean find132pattern(int[] nums) {
int n = nums.length;
int last = Integer.MIN_VALUE; // 132中的2
Stack<Integer> sta = new Stack<>();// 用来存储132中的3
if(nums.length < 3)
return false;
for(int i=n-1; i>=0; i--){
if(nums[i] < last) // 若出现132中的1则返回正确值
return true;
// 若当前值大于或等于2则更新2（2为栈中小于当前值的最大元素）
while(!sta.isEmpty() && sta.peek() < nums[i]){
last = sta.pop();
}
// 将当前值压入栈中
sta.push(nums[i]);
}
return false;
}