<div id="readability-page-1" class="page"><section><section><h3 id="category">알고리즘 / Solution</h3><ul><li> akasai
 </li><li>10월 20 화, 2020
 · 1 min read
 </li></ul><section data-v-fc5a9040=""><ul data-v-fc5a9040=""><li data-v-fc5a9040=""><a href="/tag/Algorithm/" data-v-fc5a9040="">
 Algorithm
 </a></li><li data-v-fc5a9040=""><a href="/tag/Leetcode/" data-v-fc5a9040="">
 Leetcode
 </a></li><li data-v-fc5a9040=""><a href="/tag/Typescript/" data-v-fc5a9040="">
 Typescript
 </a></li></ul></section></section><ins data-ad-client="ca-pub-7791595479585064" data-ad-slot="1631172523" data-ad-format="auto" data-full-width-responsive="true"></ins><section><article><img src="https://raw.githubusercontent.com/akasai/Algorithm-Solutions/master/Leetcode/leetcode-logo.png">
 
<blockquote>
<a href="https://leetcode.com/problems/majority-element" target="_blank" rel="nofollow noopener noreferrer">String Compression</a>
</blockquote>
<hr>
<h2 id="how-to-solve">How To Solve</h2>
처음에는 2중 반복문을 생각했다.
다시 생각해보니 카운팅류의 문제는 <code>Hash</code>를 사용하는게 좋을 것 같아서
다시 구성했다. 
Solution을 찾아보니 분할정복으로도 해결할 수 있다고 하는데...
 
<h2 id="solution">Solution</h2>
<pre><code>function majorityElement(nums: number[]): number {
 const pivot = Math.floor(nums.length / 2)

 const countMap = nums.reduce((obj: any, v: number) =&gt; {
 if (obj[v]) obj[v]++
 else obj[v] = 1
 return obj
 }, {})

 let answer = 0
 for (const key in countMap) {
 if (countMap[key] &gt; pivot) {
 answer = +key
 break
 }
 }
 return answer
}</code></pre>
<hr>
 
<h2 id="reference">Reference</h2>

<a href="https://github.com/akasai/Algorithm-Solutions/blob/master/Leetcode/Solution/32.Majority_Element.ts" target="_blank" rel="nofollow noopener noreferrer">Github</a>

</article></section><ins data-ad-client="ca-pub-7791595479585064" data-ad-slot="6759499833" data-ad-format="auto" data-full-width-responsive="true"></ins></section></div>

devlog.akasai

akasai

![](https://raw.githubusercontent.com/akasai/Algorithm-Solutions/master/Leetcode/leetcode-logo.png)
> [String Compression](https://leetcode.com/problems/majority-element)
***
## How To Solve
처음에는 2중 반복문을 생각했다.
다시 생각해보니 카운팅류의 문제는 `Hash`를 사용하는게 좋을 것 같아서
다시 구성했다. 
Solution을 찾아보니 **분할정복**으로도 해결할 수 있다고 하는데...
## Solution
```typescript
function majorityElement(nums: number[]): number {
  const pivot = Math.floor(nums.length / 2)
  const countMap = nums.reduce((obj: any, v: number) => {
    if (obj[v]) obj[v]++
    else obj[v] = 1
    return obj
  }, {})
  let answer = 0
  for (const key in countMap) {
    if (countMap[key] > pivot) {
      answer = +key
      break
    }
  }
  return answer
}
```
***
## Reference
[Github](https://github.com/akasai/Algorithm-Solutions/blob/master/Leetcode/Solution/32.Majority_Element.ts)

[Leetcode] Majority Element

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