Majority Element （算法）

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

public class Solution {
    public int majorityElement(int[] nums) {
        //way1:genuis
        int major=nums[0], count = 1;
        for(int i=1; i<nums.length;i++){
            if(count==0){
                count++;
                major=nums[i];
            }else if(major==nums[i]){
                count++;
            }else count--;
        }
        return major;
        
        //way2:most efficient
        int len = nums.length;
        Arrays.sort(nums);
        return nums[len/2];
        
        //way3:mine
        //n个元素的集合，必须出现(n/2+1)次以上 
        int len = nums.length;
        if(len==1)
            return nums[0];
        Arrays.sort(nums);
        int majorLen = (len/2)+1;
        int pointer = 0;
        int count = 1;
        while(pointer<len-1){
            if(count>=majorLen)
              break;
            else{
                if((pointer+1<len)&&(nums[pointer]==nums[pointer+1]))
                    count++;
                else
                    count=1;
                pointer++;
            }
        }
        return nums[pointer];
    }
}