169. Majority Element

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?

解题思路：关键在于多数元素>n/2， 多数元素唯一;

1. 先排序，因为关键，故（n）/2 肯定为指定的数

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

2. 建立hashmap。存储出现的所有的数的个数，当该数的个数>n/2时，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. 摩尔投票法。已知多数元素>n/2,那么比其他所有的元素累计还要多。利用摩尔投票类似“抵消”，或者多方打仗，如果自己人碰上自己人，加一，无伤亡；如果碰到其他人，两个人都伤亡。最后剩下来的肯定是拥有士兵最多的队伍。

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;
    }
}