题目描述
多数元素
给定一个大小为 n 的数组,找到其中的多数元素。多数元素是指在数组中出现次数 大于 ⌊ n/2 ⌋ 的元素。
你可以假设数组是非空的,并且给定的数组总是存在多数元素。
- 示例 1:
输入:[3,2,3]
输出:3 - 示例 2:
输入:[2,2,1,1,1,2,2]
输出:2
一、我的解答
拒绝O(n^2)从你我做起~
排序,前后指针,计数判断
class Solution {
public int majorityElement(int[] nums) {
int len = nums.length;
if(len == 1){
return nums[0];
}
Arrays.sort(nums);
int first,second;
int count = 1;
for(int i = 0;i<len - 1;i++){
first = nums[i];
second = nums[i+1];
if(first == second){
count++;
if(count > (len/2)){
return first;
}
}
else{
count = 1;
}
}
return 0;
}
}
二、题解
2.1哈希表
先建表,元素作为键,出现的次数作为值,后 遍历每个entry,找出最大的value对应的key
class Solution {
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 majorityElement(int[] nums) {
Map<Integer,Integer> counts = countNums(nums);
Map.Entry<Integer,Integer> majorityEntry = null;
//遍历每个entry,找出最大的value对应的key
for(Map.Entry<Integer,Integer> entry : counts.entrySet()){
if(majorityEntry == null || entry.getValue() > majorityEntry.getValue()){
majorityEntry = entry;
}
}
return majorityEntry.getKey();
}
}
2.2排序
感觉这个很巧妙
如果将数组 nums 中的所有元素按照单调递增或单调递减的顺序排序,那么下标为 n/2 向下取整
的元素(下标从 0 开始)一定是众数。
class Solution {
public int majorityElement(int[] nums) {
Arrays.sort(nums);
return nums[nums.length / 2];
}
}
2.3随机化
因为超过2/n ⌋ 的数组下标被众数占据了,这样我们随机挑选一个下标对应的元素并验证,有很大的概率能找到众数。
算法
由于一个给定的下标对应的数字很有可能是众数,我们随机挑选一个下标,检查它是否是众数,如果是就返回,否则继续随机挑选。
class Solution {
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 majorityElement(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;
}
}
}
}
2.4分治
class Solution {
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 majorityElement(int[] nums) {
return majorityElementRec(nums, 0, nums.length - 1);
}
}
2.5投票算法
class Solution {
public int majorityElement(int[] nums) {
int count = 0;
Integer candidate = null;
for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
return candidate;
}
}