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);
}