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# 334. 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.

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

**Example 1:**

```
Input: [1,2,3,4,5]
Output: true
```

**Example 2:**

```
Input: [5,4,3,2,1]
Output: false
```

这个思路是使用两个指针m1和m2，初始化为整型最大值，我们遍历数组，如果m1大于等于当前数字，则将当前数字赋给m1；如果m1小于当前数字且m2大于等于当前数字，那么将当前数字赋给m2，一旦m2被更新了，说明一定会有一个数小于m2，那么我们就成功的组成了一个长度为2的递增子序列，所以我们一旦遍历到比m2还大的数，我们直接返回ture。如果我们遇到比m1小的数，还是要更新m1，有可能的话也要更新m2为更小的值，毕竟m2的值越小，能组成长度为3的递增序列的可能性越大

```java
class Solution {
    public boolean increasingTriplet(int[] nums) {
        if(nums == null || nums.length == 0) return false;
        
        
        int m1 = Integer.MAX_VALUE, m2 = Integer.MAX_VALUE;
        
        for(int i : nums){
            if(m1 >= i) m1 = i;
            else if(m2 >= i) m2 = i;
            else return true;
        }
        
        return false;
    }
}
```