利用栈的暴力解法，O(n^2)的时间复杂度，但是leetcode报错超时。

#include <stack>

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        int m = nums.size();
        int n = 2;
        for (int i = 0; i <= m - 3; i++) {
            stack<int> stack;
            stack.push(nums[i]);

            for (int j = i+1; j < m; j++) {
                if (nums[j] > nums[i]) {
                    if (stack.size() == 1) {
                        stack.push(nums[j]);
                    } else if (nums[j] <= stack.top()) {
                        stack.pop();
                        stack.push(nums[j]);
                    } else {
                        stack.push(nums[j]);
                    }
                }

                if (stack.size() == 3) {
                    return true;
                }
            }
        }

        return false;
    }
};

#include <stack>

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        int m = nums.size();
        vector<int> left(m, 0);
        left[0] = nums[0];
        vector<int> right(m, 0);
        right[m-1] = nums[m-1]; 

        for (int i = 1; i < m; i++) {
            if (nums[i-1] < left[i-1]) {
                left[i] = nums[i-1];
            } else {
                left[i] = left[i-1];
            }
        }

        for (int i = m-2; i >= 0; i--) {
            if (nums[i+1] > right[i+1]) {
                right[i] = nums[i+1];
            } else {
                right[i] = right[i+1];
            }
        }

        for (int i = 0; i < m; i++) {
            if (left[i] < nums[i] && right[i] > nums[i]) {
                return true;
            }
        }

        return false;
    }
};