Source

Given an array of integers,
the majority number is the number that occurs more than 1/3 of the size of the array.

Find it.

Example
Given [1, 2, 1, 2, 1, 3, 3], return 1.

Note
There is only one majority number in the array.

Challenge
O(n) time and O(1) extra space.

題解

題Majority Number的升級版，之前那道題是『兩兩抵消』，這道題自然則需要『三三抵消』，不過『三三抵消』需要注意不少細節，比如兩個不同數的新增順序和新增條件。

Java

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: The majority number that occurs more than 1/3
     */
    public int majorityNumber(ArrayList<Integer> nums) {
        if (nums == null || nums.isEmpty()) return -1;

        // pair
        int key1 = -1, key2 = -1;
        
 
int count1 = 0, count2 = 0;
        for (int num : nums) {
            if (count1 == 0) {
                key1 = num;
                count1 = 1;
                continue;
            } else if (count2 == 0 && key1 != num) {
                key2 = num;
                count2 = 1;
                
 
continue;
            }
            if (key1 == num) {
                count1++;
            } else if (key2 == num) {
                count2++;
            } else {
                count1--;
                count2--;
            }
        }

        count1 = 0;
        count2 = 0;
        for (int num : nums) {
            if (key1 == num) {
                count1++;
            } else if (key2 == num) {
                count2++;
            }
        }
        return count1 > count2 ? key1 : key2;
    }
}

原始碼分析

首先處理count == 0的情況，這裡需要注意的是count2 == 0 && key1 = num, 不重不漏。最後再次遍歷原陣列也必不可少，因為由於新增順序的區別，count1 和 count2的大小隻具有相對意義，還需要最後再次比較其真實計數器值。

複雜度分析

時間複雜度 O(n), 空間複雜度 O(2×2)= O(1)O(2×2)=O(1).