LintCode 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.

n will be less than 20,000.

样例

Given nums = [1, 2, 3, 4]
return False // There is no 132 pattern in the sequence.

Given nums = [3, 1, 4, 2]
return True // There is a 132 pattern in the sequence: [1, 4, 2].

思路

蛮力搜索找出 3 个合乎条件的数复杂度是 n 的 3 次方，运行超时。优化为保存一个数之前可能形成 132 模式的几组最大最小数，判断这个数否在这个几组最大最小数之间。

class Solution {public:    /**     * @param nums a list of n integers     * @return true if there is a 132 pattern or false     */    bool find132pattern(vector<int>& nums) {        // Write your code here        if(nums.size() < 3) return false;        vector<int> stack;        stack.push_back(nums[0]); stack.push_back(nums[1]);        for(int i = 0; i < nums.size(); ++i) {            int cur = nums[i];            for(int j = 0; j < stack.size(); j += 2) {                if(cur > stack[j] && cur < stack[j + 1]) {                    return true;                }                if(cur > stack[j + 1]) {                    stack[j + 1] = cur;                }            }            if(cur < stack[stack.size() - 2]) {                stack.push_back(cur); stack.push_back(cur);            }        }        return false;    }};