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_169_MajorityElement.java
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_169_MajorityElement.java
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package leetcode_1To300;
import java.util.Arrays;
import java.util.HashMap;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _169_MajorityElement {
/**
* 169. Majority Element
* 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
* @param nums
* @return
*/
// time : O(nlogn) space : O(1)
public int majorityElement1(int[] nums) {
Arrays.sort(nums);
return nums[nums.length / 2];
}
// time : O(n) space : O(n)
public int majorityElement2(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
int res = 0;
for (int num : nums) {
if (!map.containsKey(num)) {
map.put(num, 1);
} else {
map.put(num, map.get(num) + 1);
}
if (map.get(num) > nums.length / 2) {
res = num;
break;
}
}
return res;
}
// Moore voting algorithm
// 每次都找出一对不同的元素,从数组中删掉,直到数组为空或只有一种元素。
// 不难证明,如果存在元素e出现频率超过半数,那么数组中最后剩下的就只有e。
// [1,2,3,3,3]
// time : O(n) space : O(1)
public int majorityElement3(int[] nums) {
int count = 0;
int res = 0;
for (int num : nums) {
if (count == 0) {
res = num;
}
if (num != res) {
count--;
} else count++;
}
return res;
}
}