一、题目
334. Increasing Triplet Subsequence
Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
Example 1:
Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.
Example 3:
Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
Constraints:
1 <= nums.length <= 5 * 105
-231 <= nums[i] <= 231 - 1
Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?
二、题解
O(n)的时间复杂度，INT_MAX表示int的最大值文章来源：https://www.toymoban.com/news/detail-724118.html
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
int n = nums.size();
int first = INT_MAX, 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;
}
};
还有一种O(n^2)的时间复杂度算法，通过先固定j，再对i和k进行循环。文章来源地址https://www.toymoban.com/news/detail-724118.html
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