Majority Element II

描述

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

Follow-up: Could you solve the problem in linear time and in O(1) space?

Example 1:

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

Example 2:

Input: nums = [1] > Output: [1]

Example 3:

Input: nums = [1,2] > Output: [1,2]

Constraints:

• 1 <= nums.length <= 5 * $10^4$
• -$10^9$ <= nums[i] <= $10^9$

分析

给定一个长度为n的数组，我们知道下面一些结论是成立的：

• 最多只有一个元素能出现超过 ⌊n/2⌋ 次.
• 最多只有两个元素能出现超过 ⌊n/3⌋ 次.
• 最多只有三个元素能出现超过 ⌊n/4⌋ 次.

以此类推。

Boyer-Moore 投票算法就是依据上面的原理而实现的。

代码

Python

# Majority Element II
# Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution(object):
    def majorityElement(self, nums: List[int])->List[int]:
        # 1st pass
        count1, count2 = 0, 0
        # candidates must be initalized to different values
        candidate1, candidate2 = 0, 1

        for num in nums:
            if candidate1 == num:
                count1 += 1
            elif candidate2 == num:
                count2 += 1
            elif count1 == 0:
                candidate1 = num
                count1 = 1
            elif count2 == 0:
                candidate2 = num
                count2 = 1
            else:
                count1 -= 1
                count2 -= 1

        # 2nd pass
        result = []
        for c in [candidate1, candidate2]:
            if nums.count(c) > len(nums)/3:
                result.append(c)

        return result

Java

// Majority Element II
// Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution {
    public List<Integer> majorityElement(int[] nums) {
        // 1st pass
        int count1 = 0, count2 = 0;
        // candidates must be initalized to different values
        int candidate1 = 0, candidate2 = 1;

        for (int num : nums) {
            if (candidate1 == num) {
                count1++;
            } else if (candidate2 == num) {
                count2++;
            } else if (count1 == 0) {
                candidate1 = num;
                count1 = 1;
            } else if (count2 == 0) {
                candidate2 = num;
                count2 = 1;
            } else {
                count1--;
                count2--;
            }
        }

        // 2nd pass
        List<Integer> result = new ArrayList<>();
        count1 = 0;
        count2 = 0;
        for (int num : nums) {
            if (candidate1 == num) count1++;
            if (candidate2 == num) count2++;
        }

        if (count1 > nums.length/3) result.add(candidate1);
        if (count2 > nums.length/3) result.add(candidate2);

        return result;
    }
}

C++

// Majority Element II
// Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution {
public:
    vector<int> majorityElement(const vector<int>& nums) {
        // 1st pass
        int count1 = 0, count2 = 0;
        // candidates must be initalized to different values
        int candidate1 = 0, candidate2 = 1;

        for (int num : nums) {
            if (candidate1 == num) {
                count1++;
            } else if (candidate2 == num) {
                count2++;
            } else if (count1 == 0) {
                candidate1 = num;
                count1 = 1;
            } else if (count2 == 0) {
                candidate2 = num;
                count2 = 1;
            } else {
                count1--;
                count2--;
            }
        }

        // 2nd pass
        vector<int> result;
        for (int c : vector<int>{candidate1, candidate2}) {
            if (std::count(nums.begin(), nums.end(), c) > nums.size()/3) {
                result.push_back(c);
            }
        }

        return result;
    }
};

相关题目

• Majority Element