Majority Element

1. Question

找数组中的大多数元素，即元素个数超过一般的那个元素

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.

2. Solution

排序后中位数即为所求，该问题可转化为求中位数问题。

2.1 O(nlogn)

先排序，那么中位数即为所求。

public class Solution {
    public int majorityElement(int[] num){
        Arrays.sort(num);
        return num[num.length/2];
    }
}

或者，利用快排思想，求中位数。

public class Solution {
    //similar to quicksort, every time you pick a branch to find.
    
    public int partition( int[] num, int start, int end ){
        if( num == null )    return 0;
        
        int k = num[start];
        int i = start+1;
        int j = end;
        
        while( i<=j ){
            for( ; i<=j && num[i]<=k; i++ );
            for( ; i<=j && num[j]>k; j-- );
            if( i<=j ){
                int temp = num[j];
                num[j] = num[i];
                num[i] = temp;
                i++;
                j--;
            }
            else
                break;
        }
        
        return i-1;
    }
    
    public int majorityElement( int[] num){
        int start = 0;
        int end = num.length-1;
        int medium = num.length/2;
        while( true ){
            int i = partition(num, start, end);
            if( i==medium )
                break;
            else if( i<medium )
                start = i+1;
            else if( i>medium )
                end = i-1;                
        }
        return num[medium];
    }
}

2.2 O(n)

采用hashtable，实际计算元素个数。效果不如排序好，因为操作复杂，运算量大，所以实际效果不好。

import java.util.Hashtable;
public class Solution {
        //using hashtable, O(n)
        public int majorityElement(int[] num){
            Hashtable<Integer, Integer> elements = new Hashtable<Integer, Integer>();
            int medium = num.length/2;
            for( int i=0; i<num.length; i++ ){
                if( elements.containsKey(num[i]) ){
                    int val = elements.get(num[i]).intValue()+1;
                    elements.replace(num[i], val);
                }
                else
                    elements.put(num[i], 1);

                if( elements.get(num[i]).intValue() > medium)
                    return num[i];
            }
            return -1;
        }
}