[?]*Majority Element II

標籤：

題目：

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.

 

解題思路：

與Majority Elements不同是這裡需要維護兩個變數n1和n2，如果下一個數與這兩個數都不同的時候，count1和count2都減一，如果下一個數與n1或者n2相等的時候，對應的count++。最後的結果必定在n1或者n2中。

 

代碼：

 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/

[?]*Majority Element II