LeetCode:Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

　　題乾的意思是給定未排序數組，從中找到三個滿足arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1遞增數字時返回true，找不到返回false。

解題思路：

　　設定兩個數字min和middle，min保存已發現的最小值，middle保存僅比min大的第二小值。

if num <= min 則 min = num
else if num <= middle 則 middle = num
else 則return true；原因是 此時的num 肯定比min,middle都大，有min < middle < num

　　源碼如下：

/**
 * Increasing Triplet Subsequence
 * @param nums
 * @return
 */
public boolean increasingTriplet(int[] nums) {
    int min = Integer.MAX_VALUE;
    int middle = Integer.MAX_VALUE;
    for (int num : nums) {
        if (num <= min) {
            min = num;
        } else if (num <= middle) {
            middle = num;
        } else {
            return true;
        }
    }
    return false;
}