12.17 Majority Number II

Question

• leetcode: Majority Element II | LeetCode OJ
• lintcode: (47) Majority Number II

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.

题解

题 Majority Number 的升级版，之前那道题是『两两抵消』，这道题自然则需要『三三抵消』，不过『三三抵消』需要注意不少细节，比如两个不同数的添加顺序和添加条件。

C++

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: The majority number occurs more than 1/3.
     */
    int majorityNumber(vector<int> nums) {
        if (nums.empty()) return -1;

        int k1 = 0, k2 = 0, c1 = 0, c2 = 0;
        for (auto n : nums) {
            if (!c1 || k1 == n) {
                k1 = n;
                c1++;
            } else if (!c2 || k2 == n) {
                k2 = n;
                c2++;
            } else {
                c1--;
                c2--;
            }
        }

        c1 = 0; 
        c2 = 0;
        for (auto n : nums) {
            if (n == k1) c1++;
            if (n == k2) c2++;
        }
        return c1 > c2 ? k1 : k2;
    }
};

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;
    }
}

源码分析

首先处理count == 0的情况，这里需要注意的是count2 == 0 && key1 = num, 不重不漏。最后再次遍历原数组也必不可少，因为由于添加顺序的区别，count1 和 count2的大小只具有相对意义，还需要最后再次比较其真实计数器值。

复杂度分析

时间复杂度 $$O(n)$$, 空间复杂度 $$O(2 \times 2) = O(1)$$.

Reference

• Majority Number II 参考程序 Java/C++/Python