Majority Element （算法）

Posted 2020-08-03 静静静流

自己写了一种方法，看了别人的解析之后觉得自己的方法好low，不高效也不明智。所以决定记录下来。

 

题目要求

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.

相关TAG

 Array　　 Divide and Conquer 　　Bit Manipulation

代码

 1 public class Solution {
 2     public int majorityElement(int[] nums) {
 3         //way1:genuis
 4         int major=nums[0], count = 1;
 5         for(int i=1; i<nums.length;i++){
 6             if(count==0){
 7                 count++;
 8                 major=nums[i];
 9             }else if(major==nums[i]){
10                 count++;
11             }else count--;
12         }
13         return major;
14         
15         //way2:most efficient
16         int len = nums.length;
17         Arrays.sort(nums);
18         return nums[len/2];
19         
20         //way3:mine
21         //n个元素的集合，必须出现(n/2+1)次以上 
22         int len = nums.length;
23         if(len==1)
24             return nums[0];
25         Arrays.sort(nums);
26         int majorLen = (len/2)+1;
27         int pointer = 0;
28         int count = 1;
29         while(pointer<len-1){
30             if(count>=majorLen)
31               break;
32             else{
33                 if((pointer+1<len)&&(nums[pointer]==nums[pointer+1]))
34                     count++;
35                 else
36                     count=1;
37                 pointer++;
38             }
39         }
40         return nums[pointer];
41     }
42 }