169. Majority Element

难度: Easy

题目描述

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
• -109 <= nums[i] <= 109
• The input is generated such that a majority element will exist in the array.

 

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

解题思路

代码实现

解决方案

java

class Solution {
    public int majorityElement(int[] nums) {
        //假设nums[0]是多数元素，用count=1 记录多数元素的数量
        //遍历后面的元素，如果和多数元素相同 count++;如果不同count--;
        //如果count==0时，重置多数元素为后一个元素，count==1;遍历完成后就可获得多数元素

        int moreEle = nums[0];
        int count = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == moreEle) {
                count++;
            } else {
                count--;
            }
            if (count == 0) {
                moreEle = nums[i + 1];
                count = 1;
                i++;
            }
        }

        return moreEle;

    }
}