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;
     }
 }