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                # 0334. 遞增的三元子序列 ## 題目地址(334. 遞增的三元子序列) <https://leetcode-cn.com/problems/increasing-triplet-subsequence/> ## 題目描述 ``` <pre class="calibre18">``` 給定一個未排序的數組,判斷這個數組中是否存在長度為 3 的遞增子序列。 數學表達式如下: 如果存在這樣的 i, j, k, 且滿足 0 ≤ i < j < k ≤ n-1, 使得 arr[i] < arr[j] < arr[k] ,返回 true ; 否則返回 false 。 說明: 要求算法的時間復雜度為 O(n),空間復雜度為 O(1) 。 示例 1: 輸入: [1,2,3,4,5] 輸出: true 示例 2: 輸入: [5,4,3,2,1] 輸出: false ``` ``` ## 前置知識 - 雙指針 ## 公司 - 百度 - 字節 ## 思路 這道題是求解順序數字是否有三個遞增的排列, 注意這里沒有要求連續的,因此諸如滑動窗口的思路是不可以的。 題目要求O(n)的時間復雜度和O(1)的空間復雜度,因此暴力的做法就不用考慮了。 我們的目標就是`依次`找到三個數字,其順序是遞增的。因此我們的做法可以是依次遍歷, 然后維護三個變量,分別記錄最小值,第二小值,第三小值。只要我們能夠填滿這三個變量就返回true,否則返回false。 ![](https://img.kancloud.cn/a1/29/a129a64c6a96b139bc4c6080bd6c2e5d_831x697.jpg) ## 關鍵點解析 - 維護三個變量,分別記錄最小值,第二小值,第三小值。只要我們能夠填滿這三個變量就返回true,否則返回false ## 代碼 ``` <pre class="calibre18">``` <span class="hljs-title">/* * @lc app=leetcode id=334 lang=javascript * * [334] Increasing Triplet Subsequence * * https://leetcode.com/problems/increasing-triplet-subsequence/description/ * * algorithms * Medium (39.47%) * Total Accepted: 89.6K * Total Submissions: 226.6K * Testcase Example: '[1,2,3,4,5]' * * Given an unsorted array return whether an increasing subsequence of length 3 * exists or not in the array. * * Formally the function should: * * Return true if there exists i, j, k * such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return * false. * * Note: Your algorithm should run in O(n) time complexity and O(1) space * complexity. * * * Example 1: * * * Input: [1,2,3,4,5] * Output: true * * * * Example 2: * * * Input: [5,4,3,2,1] * Output: false * * * */</span> <span class="hljs-title">/** * @param {number[]} nums * @return {boolean} */</span> <span class="hljs-keyword">var</span> increasingTriplet = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">nums</span>) </span>{ <span class="hljs-keyword">if</span> (nums.length < <span class="hljs-params">3</span>) <span class="hljs-keyword">return</span> <span class="hljs-params">false</span>; <span class="hljs-keyword">let</span> n1 = <span class="hljs-params">Number</span>.MAX_VALUE; <span class="hljs-keyword">let</span> n2 = <span class="hljs-params">Number</span>.MAX_VALUE; <span class="hljs-keyword">for</span>(<span class="hljs-keyword">let</span> i = <span class="hljs-params">0</span>; i < nums.length; i++) { <span class="hljs-keyword">if</span> (nums[i] <= n1) { n1 = nums[i] } <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (nums[i] <= n2) { n2 = nums[i] } <span class="hljs-keyword">else</span> { <span class="hljs-keyword">return</span> <span class="hljs-params">true</span>; } } <span class="hljs-keyword">return</span> <span class="hljs-params">false</span>; }; ``` ``` **復雜度分析** - 時間復雜度:O(N)O(N)O(N) - 空間復雜度:O(1)O(1)O(1) 大家對此有何看法,歡迎給我留言,我有時間都會一一查看回答。更多算法套路可以訪問我的 LeetCode 題解倉庫:<https://github.com/azl397985856/leetcode> 。 目前已經 37K star 啦。 大家也可以關注我的公眾號《力扣加加》帶你啃下算法這塊硬骨頭。 ![](https://img.kancloud.cn/cf/0f/cf0fc0dd21e94b443dd8bca6cc15b34b_900x500.jpg)
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