LeetCode 169 : Majority Element

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.

自己的解法：先排序, length/2 处的元素就是 majority

时间复杂度 : O(nlogn)

public class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length/2];
    }
}

另外一种比较容易想到的解法：运用hash table

时间复杂度：O(n)

空间复杂度 : O(n)

public class Solution {
    public int majorityElement(int[] nums) {
        HashMap<Integer, Integer> map = new HashMap<>();
        int key, val, max = 1, maxKey = nums[0];
        for (int i = 0; i < nums.length; i++) {
            key = nums[i];
            if (!map.containsKey(key))
                map.put(key, 1);
            else {
                val = map.get(key) + 1;
                if (max < val) {
                    max = val;
                    maxKey = key;
                }
                map.put(key, val);
            }
        }
        return maxKey;
    }
}

别人家的解法：

据说叫做 Moore voting algorithm

时间复杂度： O(n)

public class Solution {
    public int majorityElement(int[] nums) {
        int maj = nums[0];
        for (int i = 0, count = 0; i < nums.length; i++) {
            if (count == 0) {
                maj = nums[i];
                count++;
            } else {
                count = maj == nums[i] ? count + 1 : count - 1;
            }
        }
        return maj;
    }
}

还有一个用  Divide and Conquer 的方法

时间复杂度：O(nlogn)

public class Solution {
    public int majorityElement(int[] nums) {
        return majHelper(nums, 0, nums.length-1);
    }
    
    public int majHelper(int[] nums, int lo, int hi) {
        if (lo == hi)
            return nums[lo];
        int temp1 = majHelper(nums, lo, (lo+hi)/2);
        int temp2 = majHelper(nums, (lo+hi)/2+1, hi);
        if (temp1 == temp2)
            return temp1;
        int count1 = 0, count2 = 0;
        for (int i = lo; i <= hi; i++) {
            if (nums[i] == temp1)
                count1++;
            else if (nums[i] == temp2)
                count2++;
        }
        return count1 > count2 ? temp1 : temp2; // 这里count1，count2随便，原因是什么需仔细思考
    }
}

再补充一个利用随机的方法

worst case running time： 无穷

import java.util.Random;

public class Solution {
    public int majorityElement(int[] nums) {
        Random random = new Random();
        for (int i = 0, k; i < 30; i++) {
            k = nums[random.nextInt(nums.length)];
            for (int j = 0, count = 0; j < nums.length; j++) {
                if (k == nums[j]) count++;
                if (count > nums.length / 2) return k;
            }
        }
        return -1;
    }
}

2015-10-21