[?]*Majority Element II

题目：

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：

与Majority Elements不同是这里需要维护两个变量n1和n2，如果下一个数与这两个数都不同的时候，count1和count2都减一，如果下一个数与n1或者n2相等的时候，对应的count++。最后的结果必定在n1或者n2中。

 

代码：

 1 public class Solution {

 2     public int majorityNumber(ArrayList<Integer> nums) {

 3         int n1 = nums.get(0), count1 = 1;

 4         int index = 1;

 5         for (; index < nums.size() && nums.get(index) == n1; index++) {

 6             count1++;

 7         }

 8         if (index == nums.size()) {

 9             return n1;

10         }

11         int n2 = nums.get(index), count2 = 1;

12         for (int i = index + 1; i < nums.size(); i++) {

13             if (nums.get(i) == n1) {

14                 count1++;

15             } else if (nums.get(i) == n2) {

16                 count2++;

17             } else {

18                 if (count1 == 0) {

19                     n1 = nums.get(i);

20                     count1 = 1;

21                 } else if (count2 == 0) {

22                     n2 = nums.get(i);

23                     count2 = 1;

24                 } else {

25                     count1--;

26                     count2--;

27                 }

28             }

29         }

30         if (count1 == 0) {

31             return n2;

32         } else if(count2 == 0) {

33             return n1;

34         } else {

35             int count = 0;

36             for (int n : nums) {

37                 if (n == n1) {

38                     count++;

39                 }

40             }

41             if (count > nums.size()/3) {

42                 return n1;

43             }

44             return n2;

45         }

46     }

47 }

 

reference：https://codesolutiony.wordpress.com/2015/03/15/lintcode-majority-element-ii/