[LeetCode] 229. Majority Element II 多数元素 II

Posted 2021-01-09 轻风舞动

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]

 

169. Majority Element 的拓展，这题要求的是出现次数大于n/3的元素，并且限定了时间和空间复杂度，因此不能排序，不能使用哈希表。

解法：Boyer-Moore多数投票算法 Boyer–Moore majority vote algorithm，T：O(n)  S: O(1) 摩尔投票法 Moore Voting

Java:

public List<Integer> majorityElement(int[] nums) {
	if (nums == null || nums.length == 0)
		return new ArrayList<Integer>();
	List<Integer> result = new ArrayList<Integer>();
	int number1 = nums[0], number2 = nums[0], count1 = 0, count2 = 0, len = nums.length;
	for (int i = 0; i < len; i++) {
		if (nums[i] == number1)
			count1++;
		else if (nums[i] == number2)
			count2++;
		else if (count1 == 0) {
			number1 = nums[i];
			count1 = 1;
		} else if (count2 == 0) {
			number2 = nums[i];
			count2 = 1;
		} else {
			count1--;
			count2--;
		}
	}
	count1 = 0;
	count2 = 0;
	for (int i = 0; i < len; i++) {
		if (nums[i] == number1)
			count1++;
		else if (nums[i] == number2)
			count2++;
	}
	if (count1 > len / 3)
		result.add(number1);
	if (count2 > len / 3)
		result.add(number2);
	return result;
}　　

Python:

class Solution:
# @param {integer[]} nums
# @return {integer[]}
def majorityElement(self, nums):
    if not nums:
        return []
    count1, count2, candidate1, candidate2 = 0, 0, 0, 1
    for n in nums:
        if n == candidate1:
            count1 += 1
        elif n == candidate2:
            count2 += 1
        elif count1 == 0:
            candidate1, count1 = n, 1
        elif count2 == 0:
            candidate2, count2 = n, 1
        else:
            count1, count2 = count1 - 1, count2 - 1
    return [n for n in (candidate1, candidate2)
                    if nums.count(n) > len(nums) // 3]

Python:

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        k, n, cnts = 3, len(nums), collections.defaultdict(int)

        for i in nums:
            cnts[i] += 1
            # Detecting k items in cnts, at least one of them must have exactly
            # one in it. We will discard those k items by one for each.
            # This action keeps the same mojority numbers in the remaining numbers.
            # Because if x / n  > 1 / k is true, then (x - 1) / (n - k) > 1 / k is also true.
            if len(cnts) == k:
                for j in cnts.keys():
                    cnts[j] -= 1
                    if cnts[j] == 0:
                        del cnts[j]

        # Resets cnts for the following counting.
        for i in cnts.keys():
            cnts[i] = 0

        # Counts the occurrence of each candidate integer.
        for i in nums:
            if i in cnts:
                cnts[i] += 1

        # Selects the integer which occurs > [n / k] times.
        result = []
        for i in cnts.keys():
            if cnts[i] > n / k:
                result.append(i)

        return result

    def majorityElement2(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        return [i[0] for i in collections.Counter(nums).items() if i[1] > len(nums) / 3]　　

C++:

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        vector<int> res;
        int m = 0, n = 0, cm = 0, cn = 0;
        for (auto &a : nums) {
            if (a == m) ++cm;
            else if (a ==n) ++cn;
            else if (cm == 0) m = a, cm = 1;
            else if (cn == 0) n = a, cn = 1;
            else --cm, --cn;
        }
        cm = cn = 0;
        for (auto &a : nums) {
            if (a == m) ++cm;
            else if (a == n) ++cn;
        }
        if (cm > nums.size() / 3) res.push_back(m);
        if (cn > nums.size() / 3) res.push_back(n);
        return res;
    }
};

C++:

vector<int> majorityElement(vector<int>& nums) {
    int cnt1 = 0, cnt2 = 0, a=0, b=1;
    
    for(auto n: nums){
        if (a==n){
            cnt1++;
        }
        else if (b==n){
            cnt2++;
        }
        else if (cnt1==0){
            a = n;
            cnt1 = 1;
        }
        else if (cnt2 == 0){
            b = n;
            cnt2 = 1;
        }
        else{
            cnt1--;
            cnt2--;
        }
    }
    
    cnt1 = cnt2 = 0;
    for(auto n: nums){
        if (n==a)   cnt1++;
        else if (n==b)  cnt2++;
    }
    
    vector<int> res;
    if (cnt1 > nums.size()/3)   res.push_back(a);
    if (cnt2 > nums.size()/3)   res.push_back(b);
    return res;
}

　　

　　

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