[leedcode 229] Majority Element II

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.

public class Solution {public List<Integer> majorityElement(int[] nums) {/*首先，可以通过分析得到结论：满足条件的数字个数cnt最多为2。 证明： ifcnt>2⇒cnt× (⌊n/3⌋+1 )>n 超出原数组的大小。 然后，借鉴在数组中求出现次数超过一半的数这道题的思路： 1). 第一遍扫描，设两个计数器和变量记录数组nums[]中出现频率最高的数。 2). 第二遍扫描，计算着两个数出现的次数。 3). 判断这两个数是否符合要求，符合则存入结果集。需要注意的是：第一个for循环的判断顺序，首先要判断是否等于m或n*/List<Integer> res=new ArrayList<Integer>();if(nums==null||nums.length<=0) return res;int m=0;int n=0;int countm=0;int countn=0;for(int i=0;i<nums.length;i++){/*if(m==nums[i]) countm++;else if(n==nums[i]) countn++;else if(countm==0){countm=1;m=nums[i];}else if(countn==0){countn=1;n=nums[i];}else {countm--;countn--;}*/if(m==nums[i]||countm==0){countm++; m=nums[i];}else if(n==nums[i]||countn==0){countn++; n=nums[i];}else{countm--;countn--;}}countm=0;countn=0;for(int i=0;i<nums.length;i++){if(nums[i]==m) countm++;else if(nums[i]==n) countn++;}if(countm>nums.length/3) res.add(m);if(countn>nums.length/3) res.add(n);return res;}
}

转载于:https://www.cnblogs.com/qiaomu/p/4713306.html

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