README_EN.md

169. Majority Element

中文文档

Description

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?

Solutions

Python3

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        cnt = major = 0
        for num in nums:
            if cnt == 0:
                major = num
                cnt = 1
            else:
                cnt += (1 if major == num else -1)
        return major

Java

class Solution {
    public int majorityElement(int[] nums) {
        int cnt = 0, major = 0;
        for (int num : nums) {
            if (cnt == 0) {
                major = num;
                cnt = 1;
            } else {
                cnt += (major == num ? 1 : -1);
            }
        }
        return major;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var majorityElement = function(nums) {
    let cnt = 0;
    let major = 0;
    for (const num of nums) {
        if (cnt == 0) {
            major = num;
            cnt = 1;
        } else {
            cnt += (major == num ? 1 : -1);
        }
    }
    return major;
};

C++

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int cnt = 0, major = 0;
        for (int num : nums) {
            if (cnt == 0) {
                major = num;
                cnt = 1;
            } else {
                cnt += (major == num ? 1 : -1);
            }
        }
        return major;
    }
};

C#

public class Solution {
    public int MajorityElement(int[] nums) {
        int cnt = 0, major = 0;
        foreach (int num in nums)
        {
            if (cnt == 0)
            {
                major = num;
                cnt = 1;
            }
            else
            {
                cnt += (major == num ? 1 : -1);
            }
        }
        return major;
    }
}

Go

func majorityElement(nums []int) int {
    var cnt, major int
    for _, num := range nums {
        if cnt == 0 {
            major = num
            cnt = 1
        } else {
            if major == num {
                cnt++
            } else {
                cnt--
            }
        }
    }
    return major
}

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