add prob #456; O(N) in time and O(N) in space;

Author: randomwalk10

Diff for: 456_132_Pattern.cpp

@@ -0,0 +1,57 @@
+/*
+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: 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].
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/132-pattern
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+*/
+
+/* O(N) in time and O(N) in space */
+#include <vector>
+#include <stack>
+using namespace std;
+
+class Solution {
+public:
+    bool find132pattern(vector<int>& nums) {
+        stack<int> s, l;
+        int cur_min = INT_MAX;
+        int prev_min = INT_MAX;
+
+        for(int i=0; i<(int)nums.size(); ++i){
+            cur_min = min(cur_min, nums[i]);
+            while( (!s.empty())&&(s.top()<=nums[i]) ){
+                s.pop(); l.pop();
+            }
+            if(!s.empty()){
+                if(l.top()<nums[i]) return true;
+            }
+            s.push(nums[i]);
+            l.push(prev_min);
+            prev_min = cur_min;
+        }
+
+        return false;
+    }
+};