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.

代码如下：（方法一：超内存）

 public class Solution {
     public List<Integer> majorityElement(int[] nums) {
         List<Integer> list=new ArrayList<>();
         if(nums.length==0)
         return list;
         Arrays.sort(nums);
         int[] a=new int[nums[nums.length-1]+1];

         for(int i=0;i<nums.length;i++)
             a[nums[i]]++;

         for(int i=0;i<nums.length;)
         {
             if(a[nums[i]]>nums.length/3)
             {
                 list.add(nums[i]);
                 i=i+a[nums[i]]-1;
             }
         }

         return list;
     }
 }

方法二：（借鉴别人的）

 public class Solution {
     public List<Integer> majorityElement(int[] nums) {
         List<Integer> list=new ArrayList<>();
         if(nums.length==0)
         return list;
         if(nums.length==1)
         {
             list.add(nums[0]);
             return list;
         }
         int m1=nums[0];
         int m2=0;
         int c1=1;
         int c2=0;
         for(int i=1;i<nums.length;i++)
         {
             if(m1==nums[i])
             c1++;
             else if(m2==nums[i])
             c2++;
             else  if(c1==0)
             {
                 m1=nums[i];
                 c1=1;
             }
             else if(c2==0)
             {
                m2=nums[i];c2=1;
             }
             else {
                 c1--;c2--;
             }
         }
         c1 = 0; c2 = 0;
         for(int i=0; i<nums.length; i++) {
             if(m1 == nums[i]) ++c1;
             else if(m2 == nums[i]) ++c2;
         }
         if(c1>nums.length/3)
         list.add(m1);
         if(c2>nums.length/3)
         list.add(m2);

         return list;
     }
 }