![https://leetcode.com/problems/majority-element/]
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?
class Solution {
private static final Random RANDOM = new Random();
public int majorityElement(int[] nums) {
sort(nums);
return nums[nums.length / 2];
}
void sort(int[] nums) {
shuffle(nums);
sort(nums, 0, nums.length - 1);
}
void sort(int[] nums, int lo, int hi) {
if (hi <= lo) return;
int j = partition(nums, lo, hi);
sort(nums, lo, j - 1);
sort(nums, j + 1, hi);
}
void shuffle(int[] a) {
for (int i = 0; i < a.length; i++) {
swap(a, i, RANDOM.nextInt(a.length));
}
}
int partition(int[] a, int lo, int hi) {
int i = lo;
int j = hi + 1;
while(true) {
while (a[++i] < a[lo]) {
if (i == hi) break;
}
while (a[lo] < a[--j]) {
if (j == lo) break;
}
if (j <= i) break;
swap(a, i, j);
}
swap(a, lo, j);
return j;
}
void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}