LeetCode_169. Majority Element
2024-10-06 07:43:43
169. Majority Element
Easy
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.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
package leetcode.easy; import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Random; public class MajorityElement {
@org.junit.Test
public void test() {
int[] nums1 = { 3, 2, 3 };
int[] nums2 = { 2, 2, 1, 1, 1, 2, 2 };
System.out.println(majorityElement1(nums1));
System.out.println(majorityElement1(nums2));
System.out.println(majorityElement2(nums1));
System.out.println(majorityElement2(nums2));
System.out.println(majorityElement3(nums1));
System.out.println(majorityElement3(nums2));
System.out.println(majorityElement4(nums1));
System.out.println(majorityElement4(nums2));
System.out.println(majorityElement5(nums1));
System.out.println(majorityElement5(nums2));
System.out.println(majorityElement6(nums1));
System.out.println(majorityElement6(nums2));
} public int majorityElement1(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;
} private Map<Integer, Integer> countNums(int[] nums) {
Map<Integer, Integer> counts = new HashMap<Integer, Integer>();
for (int num : nums) {
if (!counts.containsKey(num)) {
counts.put(num, 1);
} else {
counts.put(num, counts.get(num) + 1);
}
}
return counts;
} public int majorityElement2(int[] nums) {
Map<Integer, Integer> counts = countNums(nums); Map.Entry<Integer, Integer> majorityEntry = null;
for (Map.Entry<Integer, Integer> entry : counts.entrySet()) {
if (majorityEntry == null || entry.getValue() > majorityEntry.getValue()) {
majorityEntry = entry;
}
} return majorityEntry.getKey();
} public int majorityElement3(int[] nums) {
Arrays.sort(nums);
return nums[nums.length / 2];
} private int randRange(Random rand, int min, int max) {
return rand.nextInt(max - min) + min;
} private int countOccurences(int[] nums, int num) {
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == num) {
count++;
}
}
return count;
} public int majorityElement4(int[] nums) {
Random rand = new Random(); int majorityCount = nums.length / 2; while (true) {
int candidate = nums[randRange(rand, 0, nums.length)];
if (countOccurences(nums, candidate) > majorityCount) {
return candidate;
}
}
} private int countInRange(int[] nums, int num, int lo, int hi) {
int count = 0;
for (int i = lo; i <= hi; i++) {
if (nums[i] == num) {
count++;
}
}
return count;
} private int majorityElementRec(int[] nums, int lo, int hi) {
// base case; the only element in an array of size 1 is the majority
// element.
if (lo == hi) {
return nums[lo];
} // recurse on left and right halves of this slice.
int mid = (hi - lo) / 2 + lo;
int left = majorityElementRec(nums, lo, mid);
int right = majorityElementRec(nums, mid + 1, hi); // if the two halves agree on the majority element, return it.
if (left == right) {
return left;
} // otherwise, count each element and return the "winner".
int leftCount = countInRange(nums, left, lo, hi);
int rightCount = countInRange(nums, right, lo, hi); return leftCount > rightCount ? left : right;
} public int majorityElement5(int[] nums) {
return majorityElementRec(nums, 0, nums.length - 1);
} public int majorityElement6(int[] nums) {
int count = 0;
Integer candidate = null; for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
} return candidate;
}
}
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