Increasing Triplet Subsequence(三個遞增數字子序列)

【難度：Medium】
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.
給定一亂序陣列，判斷其內部是否存在三個順序遞增的數字子序列，
如存在0<= i < j < k <= n-1，使得a[i]

解題思路

設定兩個變數fisrt，second來記錄已出現的兩個遞增的數字，遍歷陣列，且維護first<second的關係，那麼當出現一個數字大於second時，則存在符合條件的遞增子序列，返回true，否則返回false。

c++程式碼如下：

class
 
 Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        if (nums.size() < 3)
            return false;
        int first = INT_MAX;
        int second = INT_MAX;

        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] <= first) {
                first = nums[i];
            } else
 
 if (nums[i] <= second) {
                second = nums[i];
            } else {
                return true;
            }
        }
        return false;
    }
};