LeetCode | Majority Element

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.

解法一：

public class Solution {
    public int majorityElement(int[] num) {
        if(num.length == 1) return num[0];
        
        Arrays.sort(num);           //对num进行排序
        
        return num[num.length/2];   //中间值一定是majorityElement，否则不可能过半
    }
}

解法二：

public class Solution {
    public int majorityElement(int[] num) {
        if(num.length == 1) return num[0];
        
        Arrays.sort(num);                //对数组进行排序
        
        int start = num[0];
        int count = 1;
        for(int i=1; i<num.length; i++){  //遍历数组，求连续相同的元素的个数
            if(start == num[i]){          //如果与start相同，count++
                count++;
                if(count > num.length/2) return num[i];
            } else{                       //如果与start不同，从num[i]开始重新统计连续相同的个数
                start = num[i];
                count = 1;
            }
        }
        
        return -1;   //查找失败
    }
}

解法三：

public class Solution {
    public int majorityElement(int[] num) {
        if(num.length == 1) return num[0];
        
        Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();
        for(int i=0; i<num.length; i++) {    
            if(myMap.containsKey(num[i])){              //利用HashMap来统计num数组中各个元素出现的个数
                myMap.put(num[i], myMap.get(num[i])+1);   // Map<num[i], count>
            }else{
                myMap.put(num[i], 1);
            }
        }
        
        Iterator myIterator = myMap.entrySet().iterator();
        while(myIterator.hasNext()){                      //遍历Map找出count > length/2 的元素
            java.util.Map.Entry myEntry = (java.util.Map.Entry) myIterator.next();
            int count = (int) myEntry.getValue();
            int result = (int) myEntry.getKey();
            if(count > (num.length/2)) return result;
        }
        
        return -1;   //查找失败
    }
}