Find132Pattern.java

/*
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.

n will be less than 20,000.

Example
Given nums = [1, 2, 3, 4]
return False // There is no 132 pattern in the sequence.

Given nums = [3, 1, 4, 2]
return True // There is a 132 pattern in the sequence: [1, 4, 2].
 */
public class Find132Pattern {

    /*
     * @param nums: a list of n integers
     * @return: true if there is a 132 pattern or false
     */
    public boolean find132pattern(int[] nums) {
        if (nums == null) {
            return false;
        }
        Stack<Integer> stack = new Stack<Integer>();
        int ak = Integer.MIN_VALUE;
        for (int i = nums.length - 1; i >= 0; --i) {
            if (nums[i] < ak) {
                return true;
            }
            while (!stack.isEmpty() && nums[i] > stack.peek()) {
                ak = stack.pop();
            }
            stack.push(nums[i]);
        }
        return false;
    }

/*****************************************************************************/

    public boolean find132pattern(int[] nums) {
        if (nums == null || nums.length < 3) {
            return false;
        }
        int i = 0;
        while (i < nums.length) {
            for (; i < nums.length - 1 && nums[i] > nums[i + 1]; ++i);
            int j = i + 1;
            for (; j < nums.length - 1 && nums[j] < nums[j + 1]; ++j);
            for (int k = j + 1; k < nums.length; ++k) {
                if (nums[i] < nums[k] && nums[k] < nums[j]) {
                    return true;
                }
            }
            i = j + 1;
        }
        return false;
    }

}