456. 132 Pattern

Given an array of `n` integers `nums`, a 132 pattern is a subsequence of three integers `nums[i]`, `nums[j]` and `nums[k]` such that `i < j < k` and `nums[i] < nums[k] < nums[j]`.

Return `true` if there is a 132 pattern in `nums`, otherwise, return `false`.

Follow up: The `O(n^2)` is trivial, could you come up with the `O(n logn)` or the `O(n)` solution?

Example 1:

```Input: nums = [1,2,3,4]
Output: false
Explanation: There is no 132 pattern in the sequence.
```

Example 2:

```Input: nums = [3,1,4,2]
Output: true
Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
```

Example 3:

```Input: nums = [-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].
```

Constraints:

• `n == nums.length`
• `1 <= n <= 104`
• `-109 <= nums[i] <= 109`

456. 132 Pattern
``````struct Solution;

use std::i32;

impl Solution {
fn find132pattern(nums: Vec<i32>) -> bool {
let n = nums.len();
let mut a3 = i32::MIN;
let mut stack: Vec<i32> = vec![];
for i in (0..n).rev() {
if nums[i] < a3 {
return true;
} else {
while let Some(top) = stack.pop() {
if nums[i] > top {
a3 = top;
} else {
stack.push(top);
break;
}
}
}
stack.push(nums[i]);
}
false
}
}

#[test]
fn test() {
let nums = vec![1, 2, 3, 4];
let res = false;
assert_eq!(Solution::find132pattern(nums), res);
let nums = vec![3, 1, 4, 2];
let res = true;
assert_eq!(Solution::find132pattern(nums), res);
let nums = vec![-1, 3, 2, 0];
let res = true;
assert_eq!(Solution::find132pattern(nums), res);
}
``````