[LeetCode] 334. Increasing Triplet Subsequence 递增三元子序列

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.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5]
Output: true

Example 2:

Input: [5,4,3,2,1]
Output: false

给一个非排序的数组，判断是否存在一个长度为3的递增子序列。 要求：T: O(n)  S: O(1)

解法：由于时间和空间复杂度的要求，不能用排序或者DP的方法。遍历数组，用两个变量分别记录当前的最小值和第二小的值，如果发现一个比这两个都大的数，则组成了一个长度为3的子序列，返回True。如果遍历结束，没找到返回False。

Java:

public boolean increasingTriplet(int[] nums) {
        // start with two largest values, as soon as we find a number bigger than both, while both have been updated, return true.
        int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;
        for (int n : nums) {
            if (n <= small) { small = n; } // update small if n is smaller than both
            else if (n <= big) { big = n; } // update big only if greater than small but smaller than big
            else return true; // return if you find a number bigger than both
        }
        return false;
    }

Python:

def increasingTriplet(nums):
    first = second = float('inf')
    for n in nums:
        if n <= first:
            first = n
        elif n <= second:
            second = n
        else:
            return True
    return False

C++:

bool increasingTriplet(vector<int>& nums) {
    int c1 = INT_MAX, c2 = INT_MAX;
    for (int x : nums) {
        if (x <= c1) {
            c1 = x;           // c1 is min seen so far (it's a candidate for 1st element)
        } else if (x <= c2) { // here when x > c1, i.e. x might be either c2 or c3
            c2 = x;           // x is better than the current c2, store it
        } else {              // here when we have/had c1 < c2 already and x > c2
            return true;      // the increasing subsequence of 3 elements exists
        }
    }
    return false;
}

　　

　　

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