Given an integer array of size?n, find all elements that appear more than?? n/3 ??times. The algorithm should run in linear time and in O(1) space.

Hint:

How many majority elements could it possibly have?

My code:

public List<Integer> majorityElement(int[] nums) {HashMap<Integer,Integer> map = new HashMap<>();List<Integer> list = new ArrayList<>();for(int i=0;i<nums.length;i++){if(!map.containsKey(nums[i])) map.put(nums[i],0);map.put(nums[i],map.get(nums[i])+1);}for(Integer key:map.keySet()){if(map.get(key)>nums.length/3) list.add(key);}return list;}總結(jié)：哎呀媽呀，你太棒了。不過(guò)空間復(fù)雜度是O(n)。 Majority Voting Algorithm: ? ??Here

Boyer-Moore Algorithm：設(shè)出一個(gè)candidate和一個(gè)count，candidate初值隨便設(shè)，相同則count增加，否則count減小。

public class Solution {public List<Integer> majorityElement(int[] nums) {List<Integer> result =new ArrayList<>();if (nums==null) return result;int cad1 =0, cad2 = 1, count1=0, count2 = 0;for(Integer n:nums){if(n==cad1) count1+=1;else if(n==cad2) count2+=1;else if(count1 ==0){cad1= n;count1=1;}else if(count2==0){cad2 = n;count2 =1 ;}else{count1-=1;count2-=1;}}count1=0;count2=0;for(Integer n:nums){if(n==cad1) count1++;else if(n==cad2) count2++;}if(count1>nums.length/3) result.add(cad1);if(count2>nums.length/3) result.add(cad2);return result;} }

總結(jié)

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