LeetCode 169. Majority Element

原题链接在这里：https://leetcode.com/problems/majority-element/

题目：

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

题解：

Method 1：最容易想到的就是用HashMap 计数，数值大于n/2(注意不是大于等于而是大于)，就是返回值。Time Complexity: O(n). Space: O(n).

Method 2: 用sort, 返回sort后array的中值即可. Time Complexity: O(n*logn). Space: O(1).

Method 3: 维护个最常出现值，遇到相同count++, 遇到不同count--, count为0时直接更改最常出现值为nums[i]. Time Complexity: O(n). Space: O(1).

AC Java:

public class Solution {
    public int majorityElement(int[] nums) {
        /*
        //Method 1, HashMap
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0;i<nums.length; i++){
            if(!map.containsKey(nums[i])){
                map.put(nums[i],1);
            }else{
                map.put(nums[i],map.get(nums[i])+1);
            }
        }
        
        Iterator<Integer> it = map.keySet().iterator(); //Iterate HashMap
        while(it.hasNext()){
            int keyVal = it.next();
            //There is an error here: Pay attentation, it is ">", but not ">="
            //If we have three variables [3,2,3], ">=" will also return 2, 1>=3/2
            if(map.get(keyVal) > nums.length/2){
                return keyVal;
            }
        }
        
        return Integer.MIN_VALUE;
        */
        
        /*Method 2, shortcut
        Arrays.sort(nums);
        return nums[nums.length/2];
        */
        
        //Method 3
        if(nums == null || nums.length == 0)
            return Integer.MAX_VALUE;
        int res = nums[0];
        int count = 1;
        for(int i = 1; i< nums.length; i++){
            if(res == nums[i]){
                count++;
            }else if(count == 0){
                res = nums[i];
                count = 1;
            }else{
                count--;
            }
        }
        return res;
        
    }
}

AC C++:

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        if(nums.size() == 0){
            return INT_MAX;
        }

        int res = nums[0];
        int count = 1;
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] == res){
                count++;
            }else if(count == 0){
                res = nums[i];
                count = 1;
            }else{
                count--;
            }
        }

        return res;
    }
};

跟上Majority Element II.

类似Check If a Number Is Majority Element in a Sorted Array.