算法 中等 | 48. 主元素 III

题目描述

给定一个整型数组，找到主元素，它在数组中的出现次数严格大于数组元素个数的1/k

样例1

输入: [3,1,2,3,2,3,3,4,4,4] and k=3, 
输出: 3.

样例2

输入: [1,1,2] and k=3, 
输出: 1.

java题解

public class Solution {
    public int majorityNumber(ArrayList<Integer> nums, int k) {
        // count at most k keys.
        HashMap<Integer, Integer> counters = new HashMap<Integer, Integer>();
        for (Integer i : nums) {
            if (!counters.containsKey(i)) {
                counters.put(i, 1);
            } else {
                counters.put(i, counters.get(i) + 1);
            }
            
            if (counters.size() >= k) {
                removeKey(counters);
            }
        }
        
        // corner case
        if (counters.size() == 0) {
            return Integer.MIN_VALUE;
        }
        
        // recalculate counters
        for (Integer i : counters.keySet()) {
            counters.put(i, 0);
        }
        for (Integer i : nums) {
            if (counters.containsKey(i)) {
                counters.put(i, counters.get(i) + 1);
            }
        }
        
        // find the max key
        int maxCounter = 0, maxKey = 0;
        for (Integer i : counters.keySet()) {
            if (counters.get(i) > maxCounter) {
                maxCounter = counters.get(i);
                maxKey = i;
            }
        }
        
        return maxKey;
    }
    
    private void removeKey(HashMap<Integer, Integer> counters) {
        Set<Integer> keySet = counters.keySet();
        List<Integer> removeList = new ArrayList<>();
        for (Integer key : keySet) {
            counters.put(key, counters.get(key) - 1);
            if (counters.get(key) == 0) {
                removeList.add(key);
            }
        }
        for (Integer key : removeList) {
            counters.remove(key);
        }
    }
}

C++题解

struct Candidate{
    int val;
    int count;
    Candidate(int x = 0, int y = 0): val(x), count(y){}
};

class Solution {
public:
    int majorityNumber(vector<int> nums, int k) {
        if(nums.size() == 0) return -1;
        
        int n = nums.size();
        vector<Candidate> m(k - 1, Candidate() );
        for(int i = 0; i < n; i++){
            int l;
            for(l = 0; l < k - 1; l++){
                if(m[l].val == nums[i]){
                    m[l].count++;
                    break;
                }
            }
            if(l == k - 1){
                
                for(l = 0; l < k - 1; l++){
                    if(m[l].count == 0){
                        m[l].val = nums[i];
                        m[l].count++;
                        break;
                    }
                }
                
                if(l == k - 1){
                    for(l = 0; l < k - 1; l++){
                        m[l].count--;
                    }
                }
            }
        }
        
        for(int j = 0; j < k - 1; j++){
            m[j].count = 0;
        }
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < k - 1; j++){
                if(nums[i] == m[j].val){
                    m[j].count++;
                }
                if(m[j].count > n / k) return m[j].val;
            }
        }
        
        return -1;
    }
};

python题解

class Solution:
    def majorityNumber(self, nums, k):
        counts = {}
        max = 0
        for num in nums:
            counts[num] = counts.get(num, 0) + 1
            if counts[num] > max:
                max = counts[num]
                majority = num
        return majority