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                # 0169. 多數元素 ## 題目地址(169. 多數元素) <https://leetcode-cn.com/problems/majority-element/> ## 題目描述 ``` <pre class="calibre18">``` 給定一個大小為 n 的數組,找到其中的多數元素。多數元素是指在數組中出現次數大于 ? n/2 ? 的元素。 你可以假設數組是非空的,并且給定的數組總是存在多數元素。 示例 1: 輸入: [3,2,3] 輸出: 3 示例 2: 輸入: [2,2,1,1,1,2,2] 輸出: 2 ``` ``` ## 前置知識 - 投票算法 ## 公司 - 阿里 - 騰訊 - 百度 - 字節 - adobe - zenefits ## 思路 符合直覺的做法是利用額外的空間去記錄每個元素出現的次數,并用一個單獨的變量記錄當前出現次數最多的元素。 但是這種做法空間復雜度較高,有沒有可能進行優化呢? 答案就是用"投票算法"。 投票算法的原理是通過不斷消除不同元素直到沒有不同元素,剩下的元素就是我們要找的元素。 ![](https://img.kancloud.cn/2c/4d/2c4d562aa3ecb35859eefc2bcd8b0aad_827x451.jpg) ## 關鍵點解析 - 投票算法 ## 代碼 - 語言支持:JS,Python Javascript Code: ``` <pre class="calibre18">``` <span class="hljs-keyword">var</span> majorityElement = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">nums</span>) </span>{ <span class="hljs-keyword">let</span> count = <span class="hljs-params">1</span>; <span class="hljs-keyword">let</span> majority = nums[<span class="hljs-params">0</span>]; <span class="hljs-keyword">for</span>(<span class="hljs-keyword">let</span> i = <span class="hljs-params">1</span>; i < nums.length; i++) { <span class="hljs-keyword">if</span> (count === <span class="hljs-params">0</span>) { majority = nums[i]; } <span class="hljs-keyword">if</span> (nums[i] === majority) { count ++; } <span class="hljs-keyword">else</span> { count --; } } <span class="hljs-keyword">return</span> majority; }; ``` ``` Python Code: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">majorityElement</span><span class="hljs-params">(self, nums: List[int])</span> -> int:</span> count, majority = <span class="hljs-params">1</span>, nums[<span class="hljs-params">0</span>] <span class="hljs-keyword">for</span> num <span class="hljs-keyword">in</span> nums[<span class="hljs-params">1</span>:]: <span class="hljs-keyword">if</span> count == <span class="hljs-params">0</span>: majority = num <span class="hljs-keyword">if</span> num == majority: count += <span class="hljs-params">1</span> <span class="hljs-keyword">else</span>: count -= <span class="hljs-params">1</span> <span class="hljs-keyword">return</span> majority ``` ``` **復雜度分析** - 時間復雜度:O(N)O(N)O(N),其中N為數組長度 - 空間復雜度:O(1)O(1)O(1) 更多題解可以訪問我的LeetCode題解倉庫:<https://github.com/azl397985856/leetcode> 。 目前已經37K star啦。 關注公眾號力扣加加,努力用清晰直白的語言還原解題思路,并且有大量圖解,手把手教你識別套路,高效刷題。 ![](https://img.kancloud.cn/cf/0f/cf0fc0dd21e94b443dd8bca6cc15b34b_900x500.jpg)
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