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
Example 1:
Input: [1,2,3,4,5]
Output: true
Example 2:
Input: [5,4,3,2,1]
Output: false
Note
常数个数版本单调栈,主要体会一下思路
通过观察这题的具体性质,我们发现在这里我们所谓的 "单调栈" 长度其实是固定的,就是3,等于3了直接返回就行。
因为 3 非常小,我们只需要存两个值,分别代表着单调栈的第一个和第二个位置;当我们碰到第三个位置的情况,就可以返回 true 了。
Code
class Solution {
public boolean increasingTriplet(int[] nums) {
int small = Integer.MAX_VALUE;
int big = small;
for (int num : nums) {
if (num < small) {
small = num;
} else if (num < big) {
big = num;
} else {
return true;
}
}
return false;
}
}
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