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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 a_i < a_k < a_j. 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: False
Explanation: There is no 132 pattern in the sequence.
Example 2:
Input: [3, 1, 4, 2]
Output: True
Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
Example 3:
Input: [-1, 3, 2, 0]
Output: True
Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
分析
- 我们维护一个栈和一个变量third，其中third就是第三个数字，也是pattern 132中的2，栈里面按顺序放所有大于third的数字，也是pattern 132中的3，那么我们在遍历的时候，如果当前数字小于third，即pattern 132中的1找到了，我们直接返回true即可，因为已经找到了，注意我们应该从后往前遍历数组.
- 这种用栈的题目，我觉得读者自己要手工模拟一下才知道其中的原理
代码
class Solution {
public:
bool find132pattern(vector<int>& nums) {
int third=INT_MIN;
stack<int> s;
for(int i=nums.size()-1;i>=0;i--){
if(nums[i]<third) return true;
else{
while(!s.empty()&&nums[i]>s.top()){
third=s.top();
s.pop();
}
}
s.push(nums[i]);
}
return false;
}
};