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
• -109 <= nums[i] <= 109

 

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 = m = 0
        for v in nums:
            if cnt == 0:
                m, cnt = v, 1
            else:
                cnt += 1 if m == v else -1
        return m

Java

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

JavaScript

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

C++

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

C#

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

Go

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

Rust

impl Solution {
    pub fn majority_element(nums: Vec<i32>) -> i32 {
        let mut m = 0;
        let mut cnt = 0;
        for &v in nums.iter() {
            if cnt == 0 {
                m = v;
                cnt = 1;
            } else {
                cnt += if m == v { 1 } else { -1 };
            }
        }
        m
    }
}

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