题目来源于 LeetCode 上第 169 号(Majority Element)问题,题目难度为 Easy,AC率52.6%
题目地址:https://leetcode.com/problems/majority-element/
题目描述
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
给定一个数组,数组的长度为n,找出数组中出现次数超过一半的元素
You may assume that the array is non-empty and the majority element always exist in the array.
你可以假设数组不为空,且元素一定存在数组中
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
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题目解析
采用的是摩尔投票算法,关于什么是摩尔投票算法,可以参考知乎这篇文章,戳这里
- 定义两个变量major和count,major 表示出现次数最多的元素,count表示暂时无法删除的元素个数
- 假设数组第一个数为出现次数最多的元素,major = nums[0],count=1
- 如果后面的数字相等 count+1,不相等 count-1
- 如果count为0,修改major为当前数,并重置 count=1
算法效率如下:
代码实现
class Solution {
public int majorityElement(int[] nums) {
int major = nums[0];
int count = 1;
for (int i = 1; i < nums.length; i++) {
if (count == 0) {
count = 1;
major = nums[i];
} else if (major == nums[i]) {
count++;
} else {
count--;
}
}
return major;
}
}
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