[leetcode]Majority Element II

思路：可以使用三向切分的快速排序算法，因为这种算法会把小于某个数，等于某个数，和大于某个数的分开，分成三部分。很容易判断等于某个数的全部数量，对于这种重复值问题，很有效。

public class Solution {List<Integer> ret = new ArrayList<Integer>();public List<Integer> majorityElement(int[] nums) {int len=nums.length;quickSort(nums,0,len-1);return ret;}private void quickSort(int[] nums,int low,int high){int n=nums.length;if((high-low+1) <= n/3)return;int lt=low,gt=high,i=low+1,temp=nums[low];while(i<=gt){if(nums[i]<temp){swap(nums,i++,lt++);}else if(nums[i]>temp){swap(nums,i,gt--);}else{i++;}}if(i-lt > n/3){ret.add(nums[lt]);}quickSort(nums,low,lt-1);quickSort(nums,i,high);}private void swap(int[] nums,int i,int j){int temp=nums[i];nums[i]=nums[j];nums[j]=temp;        }
}

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