Majority Element 解答

Solution 1

Naive way

First, sort the array using Arrays.sort in Java. Than, scan once to find the majority element. Time complexity O(nlog(n))

public class Solution {
    public int majorityElement(int[] nums) {
        int length = nums.length;
        if (length == 1)
            return nums[0];
        Arrays.sort(nums);
        int prev = nums[0];
        int count = 1;
        for (int i = 1; i < length; i++) {
            if (nums[i] == prev) {
                count++;
                if (count > length / 2) return nums[i];
            } else {
                prev = nums[i];
                count = 1;
            }
        }
        return 0;
    }
}

Solution 2

Since the majority always take more than a half space, the middle element is guaranteed to be the majority. 

public class Solution {
    public int majorityElement(int[] nums) {
        int length = nums.length;
        if (length == 1)
            return nums[0];
        Arrays.sort(nums);
        return nums[length / 2];
    }
}

Solution 3 Moore voting algorithm

As we iterate the array, we look at the current element x:

1. If the counter is 0, we set the current candidate to x and the counter to 1.
2. If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.

After one pass, the current candidate is the majority element. Runtime complexity = O(n).

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

Solution 4 Bit Manipulation

We would need 32 iterations, each calculating the number of 1's for the ith bit of all n numbers. Since a majority must exist, therefore, either count of 1's > count of 0's or vice versa (but can never be equal). The majority number’s ith bit must be the one bit that has the greater count.

Time complexity: 32 * n = T(n)

public class Solution {
    public int majorityElement(int[] nums) {
        int length = nums.length;
        if (length == 1)
            return nums[0];
        int[] dig = new int[32];
        for (int i = 0; i < length; i++) {
            int tmp = nums[i];
            for (int j = 0; j < 32; j++) {
                dig[j] += tmp & 1;
                tmp = tmp >> 1;
            }
        }
        int max = 0;
        int tmp = 1;
        for (int i = 0; i < 32; i++) {
            if (dig[i] > length / 2) {
                max = max | tmp;
            }
            tmp = tmp << 1;
        }
        return max;
    }
}