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文章觀看次數: 670
LeetCode題目翻譯 英文原文如下
Given an array nums of size n , return the majority element.
The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.
中文翻譯
給予一個長度為 n 的陣列 nums,回傳最常出現的元素(值)
最常出現的元素是一個出現超過 [n/2] 次的元素,你可以認定每個陣列中一定會存在這個元素
來看下範例
這題的意思與範例都蠻容易理解的
範例1中, 2出現了一次,3則出現了兩次, nums.Length則為3 ,故答案為3 ( 2次 >3 ÷2 )
範例2中, 2出現了四次,1則出現了三次, nums.Length則為7 ,故答案為2 ( 4次 >7 ÷2)
解題思路 第一個方法也比較直覺,迴圈遞迴進去數每個元素各自出現幾次,出現超過 [n/2] 次就回傳
那用C#來實作看看
C# 解決方案 方案1 – 簡單遞迴 + Dictionary
public class Solution {
public int MajorityElement(int[] nums) {
if(nums.Length == 1) //陣列只有一個元素就直接回傳
{
return nums[0];
}
//用來存放每個元素出現幾次的紀錄,Key為元素(數字)、Value為出現次數
Dictionary<int, int> dict = new Dictionary<int, int>();
int tmp = 0;
int check = nums.Length/2;
foreach(int num in nums)
{
if(dict.ContainsKey(num)) //如果已存在字典內,則幫現在數字加1
{
tmp = dict[num]+1;
if(tmp>check) //要是出現的次數超過[n/2]次,則回傳結果
{
return num;
}
dict[num] = tmp;
}
else
{
dict.Add(num,1); //若不存在字典內,幫字典新增
}
}
return -1;
}
}
實作上選擇用字典以Key Value的方式來查找跟儲存,應該算是比較容易理解的寫法
⭐方案2 – 很簡短的寫法
結果在寫好的兩個禮拜後,筆者在回顧曾經寫過的程式時,突然靈光一閃
既然說明中已經表示「 最常出現的元素是一個出現超過 [n/2] 次的元素」
那是不是只要經過排序,中間的那個值就會是答案呢?
舉例來說 :
1121222 ➔ 排序後為 1112222➔ 共7個元素 ➔取第四個元素(index 3)就是答案 : 2
因為那個元素佔了超過一半的長度,所以正中間的元素一定就會是那個元素
程式碼如下 :
public class Solution {
public int MajorityElement(int[] nums) {
Array.Sort(nums); //1.先對陣列進行排列
return(nums[nums.Length/2]); //2.直接取index為[nums.Length/2]的元素回傳
}
}
沒錯,就是這麼的簡潔!
結論這題比較沒什麼太高的難度,主要是答案2要換位思考一下才能理解而已
這次分享個馬克·吐溫的名言,剛好也跟多數有關
"Whenever you find yourself on the side of the majority ,
it is time to pause and reflect ."
Mark Twain 當你發現自己跟多數人站在同一側時,是時候停下來反思一下了
題目連結 : Majority Element – LeetCode
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