Description:
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.
Solution:
模拟LIS的做法,不过只需要记录最小的两个就好。
因为是递增,所以在符合
if (nums[i] > min_1 && nums[i] < min_2)
情况下,再给第二小的赋值
min_2 = nums[i];
Code:
public class Solution {
public boolean increasingTriplet(int[] nums) {
int len = nums.length;
if (len <= 2)
return false;
int min_1 = Integer.MAX_VALUE;
int min_2 = Integer.MAX_VALUE;
for (int i = 0; i < len; i++) {
if (nums[i] > min_2) {
return true;
}
if (nums[i] < min_1) {
min_1 = nums[i];
} else if (nums[i] > min_1 && nums[i] < min_2) {
min_2 = nums[i];
}
}
return false;
}
}