Increasing Triplet Subsequence
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 [1, 2, 3, 4, 5]
,
return true
.
Given [5, 4, 3, 2, 1]
,
return false
.
Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.
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class Solution { public: bool increasingTriplet(vector<int>& nums) { int i = nums.size() - 1; if (i < 2) { return false; } int y = INT_MAX; int z = INT_MAX; for (; i>0; i--) { if (nums[i] > nums[i-1]) { if (z == INT_MAX) { z = nums[i]; y = nums[i - 1]; } else if (nums[i - 1] >= z) { z = nums[i]; y = nums[i - 1]; } else if (nums[i - 1] > y && nums[i] > z) { z = nums[i]; y = nums[i - 1]; } else if (nums[i - 1] > y && nums[i] < z) { return true; }else if (nums[i - 1] == y && nums[i] < z){ return true; } else if (nums[i - 1] < y) { return true; } } } return false; } };
自己写的太复杂了,看了别人的,凝练一下,
class Solution { public: bool increasingTriplet(vector<int>& nums) { int n = nums.size(); int a = INT_MAX; int b = INT_MAX; for (int i = 0; i < n; i++) { if (nums[i] <= a) { a = nums[i]; } else if (nums[i] <= b) { b = nums[i]; } else { return true; } } return false; } };
posted on 2016-02-22 17:23 walkwalkwalk 阅读(184) 评论(0) 编辑 收藏 举报