456 132 Pattern

456. 132 Pattern

1. Question

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:
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Input: [1, 2, 3, 4]
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Output: False
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Explanation: There is no 132 pattern in the sequence.
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Example 2:
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Input: [3, 1, 4, 2]
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Output: True
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Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
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Example 3:
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Input: [-1, 3, 2, 0]
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Output: True
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Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
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2. Implementation

(1) Monotone Stack
思路: 这题只要返回true or false,所以我们从后往前遍历数组,利用单调递减stack寻找第二大的数,一旦遍历的过程中找到一个前面的数小于第二大的数,则我们找到pattern
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class Solution {
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public boolean find132pattern(int[] nums) {
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int secondMax = Integer.MIN_VALUE;
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Stack<Integer> stack = new Stack<>();
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for (int i = nums.length - 1; i >= 0; i--) {
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if (nums[i] < secondMax) {
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return true;
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}
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else {
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while (!stack.isEmpty() && nums[i] > stack.peek()) {
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secondMax = stack.pop();
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}
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stack.push(nums[i]);
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}
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}
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return false;
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}
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}
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3. Time & Space Complexity

时间和空间复杂度都是O(n)