
## 题目地址
https://leetcode.com/problems/majority-element/description/

## 题目描述

```
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example 1:

Input: [3,2,3]
Output: 3
Example 2:

Input: [2,2,1,1,1,2,2]
Output: 2

```

## 思路

符合直觉的做法是利用额外的空间去记录每个元素出现的次数，并用一个单独的变量记录当前出现次数最多的元素。

但是这种做法空间复杂度较高，有没有可能进行优化呢？ 答案就是用"投票算法"。

投票算法的原理是通过不断消除不同元素直到没有不同元素，剩下的元素就是我们要找的元素。

![169.majority-element](../assets/problems/169.majority-element.png)

## 关键点解析

- 投票算法


## 代码

* 语言支持：JS，Python

Javascript Code:

```js
/*
 * @lc app=leetcode id=169 lang=javascript
 *
 * [169] Majority Element
 *
 * https://leetcode.com/problems/majority-element/description/
 *
 * algorithms
 * Easy (51.62%)
 * Total Accepted:    365.6K
 * Total Submissions: 702.5K
 * Testcase Example:  '[3,2,3]'
 *
 * Given an array of size n, find the majority element. The majority element is
 * the element that appears more than ⌊ n/2 ⌋ times.
 *
 * You may assume that the array is non-empty and the majority element always
 * exist in the array.
 *
 * Example 1:
 *
 *
 * Input: [3,2,3]
 * Output: 3
 *
 * Example 2:
 *
 *
 * Input: [2,2,1,1,1,2,2]
 * Output: 2
 *
 *
 */
/**
 * @param {number[]} nums
 * @return {number}
 */
var majorityElement = function(nums) {
    let count = 1;
    let majority = nums[0];
    for(let i = 1; i < nums.length; i++) {
        if (count === 0) {
            majority = nums[i];
        }
        if (nums[i] === majority) {
            count ++;
        } else {
            count --;
        }
    }
    return majority;
};
```

Python Code:

```python
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        count, majority = 1, nums[0]
        for num in nums[1:]:
            if count == 0:
                majority = num
            if num == majority:
                count += 1
            else:
                count -= 1
        return majority
```
