LeetCode - 334 - Increasing Triplet Subsequence
2017-08-02 15:41
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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.
Your algorithm should run in O(n) time complexity and O(1) space complexity.
Examples:
Given
return
Given
return
找到三个数使得满足a[i] < a[j] < a[k] && i < j < k
思维题吧?
设置一个small和big,如果小于small就替换small,如果大于small小于big,就替换big。否则,大于这两个数,说明条件成立,返回true。
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.
Your algorithm should run in O(n) time complexity and O(1) space complexity.
Examples:
Given
[1, 2, 3, 4, 5],
return
true.
Given
[5, 4, 3, 2, 1],
return
false.
找到三个数使得满足a[i] < a[j] < a[k] && i < j < k
思维题吧?
设置一个small和big,如果小于small就替换small,如果大于small小于big,就替换big。否则,大于这两个数,说明条件成立,返回true。
class Solution { public: bool increasingTriplet(vector<int>& nums) { if (nums.size() < 3) return false; int a1 = INT_MAX, a2 = INT_MAX; for (auto x: nums) { if (x <= a1) a1 = x; else if (x <= a2) a2 = x; else return true; } return false; } };
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