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.

Example 1:

Input: nums = [3,2,3]
Output: 3

Example 2:

Input: nums = [2,2,1,1,1,2,2]
Output: 2

Constraints:

n == nums.length
1 <= n <= 5 * 104
-231 <= nums[i] <= 231 - 1

Follow-up: Could you solve the problem in linear time and in O(1) space?

解題思路：關鍵在於多數元素>n/2， 多數元素唯一;

1. 先排序，因為關鍵，故（n）/2 肯定為指定的數

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

2. 建立hashmap。儲存出現的所有的數的個數，當該數的個數>n/2時，直接return

class Solution {
    public static int majorityElement(int[] nums) {
        HashMap<Integer, Integer> number = new HashMap<Integer, Integer>();
        for(int num:nums){
            if(null == number.get(num)){
                number.put(num, 1);
            }else{
                number.put(num, number.get(num)+1);
            }
            if(number.get(num)> nums.length/2){
                return num;
            }
        }
        return -1;
    }
}
 

3. 摩爾投票法。已知多數元素>n/2,那麼比其他所有的元素累計還要多。利用摩爾投票類似“抵消”，或者多方打仗，如果自己人碰上自己人，加一，無傷亡；如果碰到其他人，兩個人都傷亡。最後剩下來的肯定是擁有士兵最多的隊伍。

class Solution {
    public int majorityElement(int[] nums) {
        int vote=0;
        int result=0;
        for(int num:nums){
            if(vote==0){
                result = num;
            }
            vote += num==result ? 1: -1;
        }
        return result;
    }
}