Increasing Triplet Subsequence

1. 解析

题目大意，求解子序列是否存在一个长度为3的递增子序列。

2. 分析

涉及子序列问题，一般都是动态规划。与Longest Increasing Subsequence最大的不同在于，只要求长度为3即可，思路是一样的。subsequence[i]表示[0, i]内递增子序列的最大长度，每次向后检索，然后更新当前的最大长度即可，前面依赖的状态之前已经求解过，不过题目要求O(1)的空间复杂度，所以这种解题思路不是正确的。

class Solution {
public:bool increasingTriplet(vector<int>& nums) {vector<int> subsequence(nums.size(), 1);int res = 0;for (int i = 0; i < nums.size(); ++i){for (int j = i - 1; j >= 0; j--){if (nums[i] > nums[j] && subsequence[i] < subsequence[j] + 1){ //更新当前的状态subsequence[i] = subsequence[j] + 1;}}res = max(res, subsequence[i]);if (res >= 3) return true;}return false;}
};

3. 核心解法

参考@Grandyang的解法，思路很巧妙，即建立两个指针，left指向当前最小的数，right指向比left略微大的数，所以我们的目标就是检测是否存在大于right指针的数，left始终更新的是最小值，right更新的是稍微比left大一点的数。

例如：5 4 9 3 2 11

left = 5——> 4——>3——>2

right = 9

11 return true

class Solution {
public:bool increasingTriplet(vector<int>& nums){int left = INT_MAX, right = INT_MAX;for (int num : nums){if (left >= num) left = num; //更新left，使其始终指向最小的值else if (right >= num) right = num; //更新right，使其始终指向比left大的最小值else return true;}return false;}
};

4. 类似的题目

Longest Increasing Subsequence

Largest Divisible Subset

[1]https://www.cnblogs.com/grandyang/p/5194599.html

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