Given 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
Follow-up: Could you solve the problem in linear time and in O(1) space?
// Sorting
public int majorityElement1(int[] nums) {
Arrays.sort(nums);
return nums[nums.length/2];
}
// Hashtable
public int majorityElement2(int[] nums) {
Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();
//Hashtable<Integer, Integer> myMap = new Hashtable<Integer, Integer>();
int ret=0;
for (int num: nums) {
if (!myMap.containsKey(num))
myMap.put(num, 1);
else
myMap.put(num, myMap.get(num)+1);
if (myMap.get(num)>nums.length/2) {
ret = num;
break;
}
}
return ret;
}
// Moore voting algorithm
public int majorityElement3(int[] nums) {
int count=0, ret = 0;
for (int num: nums) {
if (count==0)
ret = num;
if (num!=ret)
count--;
else
count++;
}
return ret;
}
// Bit manipulation
public int majorityElement(int[] nums) {
int[] bit = new int[32];
for (int num: nums)
for (int i=0; i<32; i++)
if ((num>>(31-i) & 1) == 1)
bit[i]++;
int ret=0;
for (int i=0; i<32; i++) {
bit[i]=bit[i]>nums.length/2?1:0;
ret += bit[i]*(1<<(31-i));
}
return ret;
}class Solution {
public int majorityElement(int[] nums) {
int majorityCount = nums.length/2;
for (int num : nums) {
int count = 0;
for (int elem : nums) {
if (elem == num) {
count += 1;
}
}
if (count > majorityCount) {
return num;
}
}
return -1;
}
}