leetcode 456. 132 Pattern

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].

题意：判断一个数组num中是否存在132顺序的三个数，注意2比1大哦。

思路1：

用一个数组mi记录前缀最小值，从后往前遍历，当前为i，用一个set记录i之后出现（已经遍历过）的数，则如果在set中找到在(mi[i-1],num[i])之间的数就存在。复杂度O(n*lgn)

代码：

class Solution_456 {
public:
    bool find132pattern(vector& nums) {
        int n=nums.size();
        if(n<=2) return false;
        vectormi(n);
        mi[0]=nums[0];
        for (int i = 1; i ri;
        set::iterator it;
        ri.insert(nums[n-1]);
        for (int i = n-2; i >0 ; --i) {
            if(nums[i]>mi[i-1]) {
                it=ri.upper_bound(mi[i-1]);
                if(it!=ri.end()&&*it

思路2：来源于点击打开链接

从后往前遍历，用一个单调递增的栈（栈顶到栈底递增）维护32，2要尽可能大，如果当前数小于2，那么就找到了，否则更新栈。比当前数小的出栈更新2，然后将当前数入栈，复杂度O(n)

代码：

class Solution_456_On {
public:
    bool find132pattern(vector& nums) {
        int n=nums.size();
        if(n<=2) return false;
        int two=INT32_MIN;
        stacks;
        for (int i = n-1; i >=0 ; --i) {
            if(nums[i]s.top()){
                two=s.top();
                s.pop();
            }
            if(s.empty()||s.top()>nums[i]) s.push(nums[i]);
        }
        return false;
    }
};