(算法分析Week16)132 Pattern[Medium]
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456. 132 Pattern[Medium]
题目来源
Description
Given a sequence of n integers a1, a2, …, an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.
Note: n will be less than 15,000.
Example 1:
Input: [1, 2, 3, 4]Output: FalseExplanation: There is no 132 pattern in the sequence.
Example 2:
Input: [3, 1, 4, 2]Output: TrueExplanation: There is a 132 pattern in the sequence: [1, 4, 2].
Example 3:
Input: [-1, 3, 2, 0]Output: TrueExplanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
Solution
给出一个数组,问是否存在132模式。132模式题目写得很清楚就不解释了。最容易想到的就是三重循环,i, j = i+1, k = j+1这样遍历,一个判断语句就可以了,但是显而易见TLE。
可以考虑降一重循环的方法,固定j,就是最大的那个数。然后在[0, j)中找最小的(以获得更大的寻找第三个数的范围),然后在j+1往后搜索满足条件的k,不过这样还是很慢。
还可以用一个栈,维护一个第二大的数second(即132中的2),栈里面放所有大于这个数的数字,从后往前遍历,如果找到一个数字小于second,则可以直接返回true(显然)。如果遍历到的数大于栈顶元素,那么我们把栈顶的元素赋值给second,并把遍历到的数压入栈。这是因为如果second值更大,会更容易找到一个满足nums[k]小于second的条件的值。
Complexity analysis
O(n)
Code
class Solution {public: bool find132pattern(vector<int>& nums) { int second = INT_MIN; stack<int> s; for (int i = nums.size() - 1; i >= 0; --i) { if (nums[i] < second) return true; else { while (!s.empty() && nums[i] > s.top()) { second = s.top(); s.pop(); } s.push(nums[i]); } } return false; }};
Result
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