169.Majority Element:修订间差异

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创建页面,内容为“=Description= {{LeetCode |id=majority-elemen |no=169 |difficulty=Easy |category=Array |collection=Top 150 |title=Majority Element |summary=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: <syntaxhighlight lang="bash"> Input: nums = [3,2,3] Output: 3 </syntaxhighlight> Example 2: <syntaxhi…”
 
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第23行: 第23行:
Output: 2
Output: 2
</syntaxhighlight>
</syntaxhighlight>
===HashMap===
{{Submission|runtime=12ms|memory=49.68MB|rp=33.25|mp=68.75}}
<syntaxhighlight lang="bash">
class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> counts = new HashMap<>();
        float half = nums.length / 2.0f;
        for(int i = 0; i < nums.length; i++) {
            int count = counts.getOrDefault(nums[i], 0) + 1;
            if(count > half) return nums[i];
            counts.put(nums[i], count);
        }
        throw new RuntimeException("No majority element found");
    }
}
</syntaxhighlight>
[[Category:Algorithm]]
[[Category:Array]]
[[Category:LeetCode]]

2024年2月19日 (一) 14:03的版本

Description

#169 Majority Element Easy
Array Top 150
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

HashMap

Runtime 12ms
Memory 49.68MB
class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> counts = new HashMap<>();
        float half = nums.length / 2.0f;
        for(int i = 0; i < nums.length; i++) {
            int count = counts.getOrDefault(nums[i], 0) + 1;
            if(count > half) return nums[i];
            counts.put(nums[i], count);
        }
        throw new RuntimeException("No majority element found");
    }
}