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.
解题思路：
《编程之美》寻找发帖水王的原题，两次遍历，一次遍历查找可能出现次数超过nums.length/3的数字，(出现三次不同的数即抛弃)，第二次验证即可。
JAVA实现如下：
public List<Integer> majorityElement(int[] nums) {
List<Integer> list = new ArrayList<Integer>();
if (nums == null || nums.length == 0)
return list;
int left = nums[0], right=nums[0];
int count1 = 1, count2 = 0;
for (int i = 1; i < nums.length; i++) {
if (nums[i] == left)
count1++;
else if(right==left){
right=nums[i];
count2=1;
}
else if(right==nums[i])
count2++;
else if(count1==0){
left=nums[i];
count1=1;
}
else if(count2==0){
right=nums[i];
count2=1;
}
else{
count2--;
count1--;
}
}
count1=0;count2=0;
for(int i=0;i<nums.length;i++){
if(nums[i]==left)
count1++;
else if(nums[i]==right)
count2++;
}
if(count1>nums.length/3)
list.add(left);
if(count2>nums.length/3)
list.add(right);
return list;
}