标签：control c++ ber form seq one 数组 ges 一个
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;
}
[LeetCode] 334. Increasing Triplet Subsequence 递增三元子序列
标签：control c++ ber form seq one 数组 ges 一个
原文地址：https://www.cnblogs.com/lightwindy/p/10459607.html
