标签：map 题目 nums int secondCount ++ majority num 众数

169. Majority ElementGiven 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
• -231 <= nums[i] <= 231 - 1

class Solution {
    public int majorityElement(int[] nums) {
        //排序
        Arrays.sort(nums);
        //排序后找中位数肯定是那个众数，因为他出现的次数超过了1/2
        return nums[nums.length/2];
    }
}

 

Could you solve the problem in linear time and in O(1) space?

class Solution {
    public int majorityElement(int[] nums) {
        int curr = 0;
        int count = 0;
        for(int num:nums){
            if(count==0) curr = num;
            if(curr==num) count++;
            else count--;
        }
        return curr;
    }
}

 

229. Majority Element II

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 * 104
• -109 <= nums[i] <= 109

class Solution {
    public List<Integer> majorityElement(int[] nums) {
        //该题最多存在两个众数，因为每个众数出现频率要超过1/3
        int first=0,firstCount = 0;
        int second=0,secondCount = 0;
        List<Integer> result  = new ArrayList();
        //第一次遍历进行摩尔投票，筛选出两个数字
        for(int num:nums){
            if(num==first) firstCount++;
            else if(num==second) secondCount++;
            //如果哪个count为0，用当前值替换
            else if(firstCount==0){
                first = num;
                firstCount++;
            }
            else if(secondCount==0){
                second = num;
                secondCount++;
            }
            //如果两个都不是，两个都--
            else{
                firstCount--;
                secondCount--;
            }
        }
        //第二次便利，校验是否大于1/3
        firstCount=0;
        secondCount=0;
        for(int num:nums){
            if(num==first) firstCount++;
            else if(num==second) secondCount++;
        }
        //将大于1/3的数字放入结果集
        if(firstCount>nums.length/3) result.add(first);
        if(secondCount>nums.length/3) result.add(second);
        return result;
    }
}

48 · Majority Number III 

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array.

There is only one majority number in the array.

Example 1:

Input:

nums = [3,1,2,3,2,3,3,4,4,4]
k = 3

Output:

3

Explanation:

3 appeared four times, more than 10 / 3.
Example 2:

Input:

nums = [1,1,2]
k = 3

Output:

1

Explanation:

1 appeared two times, more than 3 / 3.

 

 

public class Solution {
    /**
     * @param nums: A list of integers
     * @param k: An integer
     * @return: The majority number
     */
    public int majorityNumber(List<Integer> nums, int k) {
        Map<Integer,Integer> map = new HashMap();
        for(int num:nums){
            if(map.containsKey(num)) map.put(num,map.get(num)+1);
            else{
                if(map.size()<k){
                    map.put(num,1);
                }
                else{
                    List<Integer> list  = new ArrayList();
                    for(int key:map.keySet()){
                        map.put(key,map.get(key)-1);
                        if(map.get(key)==0) list.add(key);
                    }
                    for(int ele:list){
                        map.remove(ele);
                    }
                }
            }
        }
        for(int key:map.keySet()) map.put(key,0);
        for(int num:nums){
            if(map.containsKey(num)) {
                int count = map.get(num);
                map.put(num,++count);
                if(count>nums.size()/k) return num;            
            }
        }
        return 0;   
    }
}

 

 

标签：map,题目,nums,int,secondCount,++,majority,num,众数
来源： https://www.cnblogs.com/cynrjy/p/15335579.html

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