本文主要是介绍递增的三元子序列-数组334-c++，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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;}
};
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