[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
.
解题思路
初始化时设置n1、n2为0x7fffffff。遍历数组:
- 若n小于等于n1,则n1=n;
- 否则,若n小于等于n2,则n2=n;
- 否则,返回true;
该过程不断缩小n1、n2,最后查看是否具有比n1、n2都小的数。
实现代码
// Runtime: 1 ms
public class Solution {
public boolean increasingTriplet(int[] nums) {
int n1 = 0x7fffffff;
int n2 = 0x7fffffff;
for (int n : nums) {
if (n <= n1) {
n1 = n;
} else if (n <= n2) {
n2 = n;
} else {
return true;
}
}
return false;
}
}