We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
Difficulty: 简单
Related Topics: 数组, 哈希表, 分治, 计数, 排序
给定一个大小为 n的数组 nums ,返回其中的多数元素。多数元素是指在数组中出现次数 大于 ⌊ n/2 ⌋ 的元素。
n
nums
⌊ n/2 ⌋
你可以假设数组是非空的,并且给定的数组总是存在多数元素。
示例 1:
输入:nums = [3,2,3] 输出:3
示例 2:
输入:nums = [2,2,1,1,1,2,2] 输出:2
提示:
n == nums.length
**进阶:**尝试设计时间复杂度为 O(n)、空间复杂度为 O(1) 的算法解决此问题。
Language: JavaScript
/** * @param {number[]} nums * @return {number} */ var majorityElement = function(nums) { let temp = nums[0] let times = 1 for (let i = 1; i < nums.length; i++) { if (nums[i] === temp) { times++ } else { times-- if (times === 0) { temp = nums[i + 1] times = 1 i++ } } } return temp }; // 排序 // var majorityElement = function(nums) { // nums.sort((a, b) => a - b) // return nums[Math.floor(nums.length / 2)] // }
The text was updated successfully, but these errors were encountered:
No branches or pull requests
169. 多数元素
Description
Difficulty: 简单
Related Topics: 数组, 哈希表, 分治, 计数, 排序
给定一个大小为
n
的数组nums
,返回其中的多数元素。多数元素是指在数组中出现次数 大于⌊ n/2 ⌋
的元素。你可以假设数组是非空的,并且给定的数组总是存在多数元素。
示例 1:
示例 2:
提示:
n == nums.length
**进阶:**尝试设计时间复杂度为 O(n)、空间复杂度为 O(1) 的算法解决此问题。
Solution
Language: JavaScript
The text was updated successfully, but these errors were encountered: