【LeetCode】334. Increasing Triplet Subsequence 解題報告

撰寫於2022-06-19閱讀次數

334. Increasing Triplet Subsequence / Medium

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]
IOutput: true
IExplanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

Comstraints

  • 1 <= nums.length <= 5 * 10^5
  • -2^31 <= nums[i] <= 2^31 - 1

Follow Up:

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

Solution: Linear Scan

思路

這一題只要求我們給出是否存在這樣的三元遞增子序列，所以我們可以簡單的用兩個變數 minNum1 與 minNum2來測試。
如果我們找到一個數字小於 minNum1（最小的數字），那就替換掉；
如果有找到一個數字介於兩者之間，那就將 minNum2 換成這個數字。
接下來，一旦我們找到一個數字大於 minNum2，便意味著這樣的三元序列成形了，反之則是沒有存在這樣的序列。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        int minNum = INT_MAX;
        int minNum2 = INT_MAX;
        for(const auto& num: nums) {
            if(num < minNum) {
                minNum = num;
            } 
            else if(num > minNum && num < minNum2) {
                minNum2 = num;
            }
            else if(num > minNum2) {
                return true;
            }
        }
        return false;
    }
};