456. 132 Pattern

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].




// solution 1 
class Solution {
    public boolean find132pattern(int[] nums) {
        // a brute force way is to try all possible solutions , time is n * n -1 * n -2 , roughly around n ^ 3 
        // compare all the values 
        // ai < aj , ak < aj, ai < ak 
        // 81 / 95 test cases passed. tle 
        for(int i = 0; i < nums.length - 2; i++){
            for(int j = i + 1; j < nums.length - 1; j++){
                for(int k = j + 1; k < nums.length; k++){
                    int ai = nums[i];
                    int aj = nums[j];
                    int ak = nums[k];
                    if(ai < aj && ak < aj && ai < ak) return true;
                }
            }
        }
        return false;
    }
}


// sol 2 

https://leetcode.com/problems/132-pattern/discuss/94077/Java-O(n)-solution-using-stack-in-detail-explanation


https://leetcode.com/problems/132-pattern/solution/

http://www.cnblogs.com/grandyang/p/6081984.html




思路是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。如果当前数字大于栈顶元素，那么我们按顺序将栈顶数字取出，赋值给third，然后将该数字压入栈，这样保证了栈里的元素仍然都是大于third的，我们想要的顺序依旧存在，进一步来说，栈里存放的都是可以维持second > third的second值，其中的任何一个值都是大于当前的third值，如果有更大的值进来，那就等于形成了一个更优的second > third的这样一个组合，并且这时弹出的third值比以前的third值更大，为什么要保证third值更大，因为这样才可以更容易的满足当前的值first比third值小这个条件




// passed 87 / 95 cases 



class Solution {
    public boolean find132pattern(int[] nums) {
        if(nums == null || nums.length < 3) return false;
        // stack stores all the nums bigger than lastNumber 
        Stack<Integer> stack = new Stack<>();
        int lastNumber = nums[nums.length - 1];
        for(int i = nums.length - 2; i >= 0; i--){
            // if current number is bigger than the lastNumber , push it to the stack 
            // if the stack is empty, push it to the stack , if the stack is not empty, 
            // if stack.peek() < current number, then last number = stack.peek(), stack push the current number 
            // if the stack.peek() > current number , do nothing 
            
            
            // if current number is smaller than the lastNumber , 
            // if the stack is empty, do nothing 
            // if the stack is not empty, return true;
            if(nums[i] > lastNumber){
                if(stack.isEmpty()){
                    stack.push(nums[i]);
                }else{
                    if(stack.peek() < nums[i]){
                        lastNumber = stack.pop();
                        stack.push(nums[i]);
                    }
                }
            }else if(nums[i] < lastNumber){
                if(stack.isEmpty()){
                    lastNumber = nums[i];
                }else{
                    return true;
                }
            }else{
                continue;
            }
        
        
        }
        return false;
        
        
    }
}