花花酱 LeetCode 229. Majority Element II

zxi — Sat, 14 Mar 2020 06:21:09 +0000

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

Input: [3,2,3]
Output: [3]

Example 2:

Input: [1,1,1,3,3,2,2,2]
Output: [1,2]

Solution: Boyer–Moore Voting Algorithm

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  vector majorityElement(vector& nums) {
    int n1 = 0;
    int c1 = 0;
    int n2 = 1;
    int c2 = 0;
    for (int num : nums) {
      if (num == n1) {
        ++c1;
      } else if (num == n2) {
        ++c2;
      } else if (c1 == 0) {
        n1 = num;
        c1 = 1;
      } else if (c2 == 0) {
        n2 = num;
        c2 = 1;
      } else {
        --c1;
        --c2;
      }
    }
    
    c1 = c2 = 0;
    for (int num : nums) {
      if (num == n1) ++c1;
      else if (num == n2) ++c2;
    }
    
    const int c = nums.size() / 3;
    vector ans;
    if (c1 > c) ans.push_back(n1);
    if (c2 > c) ans.push_back(n2);
    return ans;
  }
};

Python3

# Author: Huahua
class Solution:
  def majorityElement(self, nums: List[int]) -> List[int]:
    n1, c1, n2, c2 = 0, 0, 1, 0
    for num in nums:
      if num == n1: c1 += 1
      elif num == n2: c2 += 1
      elif c1 == 0: n1, c1 = num, 1
      elif c2 == 0: n2, c2 = num, 1
      else: c1, c2 = c1 - 1, c2 -1
    
    c1, c2 = 0, 0
    for num in nums:
      if num == n1: c1 += 1
      elif num == n2: c2 += 1
    
    ans = []
    if c1 > len(nums) // 3: ans.append(n1)
    if c2 > len(nums) // 3: ans.append(n2)
    return ans

Boyer–Moore Archives - Huahua's Tech Road

花花酱 LeetCode 229. Majority Element II

Solution: Boyer–Moore Voting Algorithm

C++

Python3

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