LeetCode75——Day8

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

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。