leetcode笔记:Majority Element II

一. 题目描述

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.

Hint:

How many majority elements could it possibly have?

二. 题目分析

题目大意是,给定一个大小为n的整数数组,从中找出所有出现次数超过 ⌊ n/3 ⌋ 的元素。要求算法满足线性时间复杂度和O(1)的空间复杂度。

提示:一共可能有多少个majority elements?

首先应该意识到一点: 在一个数组中出现超过三分之一次的元素,这样的元素最多只能有两个,超过两个就与命题相矛盾。该题与题目Majority Element的思路相类似,只是该题中最多可能出现两个majority elements。在这里,要求线性时间复杂度和O(1)的空间复杂度,唯有使用摩尔投票法的思想来解。

摩尔投票法的基本思想很简单,在每一轮投票过程中,从数组中找出一对不同的元素,将其从数组中删除。这样不断的删除直到无法再进行投票,如果数组为空,则没有任何元素出现的次数超过该数组长度的一半。如果只存在一种元素,那么这个元素则可能为目标元素。

三. 示例代码

// C++代码
class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        vector<int> result;
        if (nums.size() == 0) return result;
        int num1 = INT_MAX, num2 = INT_MAX;
        int time1 = 0, time2 = 0;
        for (int i = 0; i < nums.size(); ++i)
        {
            if (num1 == nums[i]) ++time1;
            else if (num2 == nums[i]) ++time2;
            else if (time1 == 0)
            {
                time1 = 1;
                num1 = nums[i];
            }
            else if (time2 == 0)
            {
                time2 = 1;
                num2 = nums[i];
            }
            else
            {
                --time1;
                --time2;
            }
        }
        time1 = 0, time2 = 0;
        for (int i = 0; i < nums.size(); ++i)
        {
            if (num1 == nums[i]) ++time1;
            else if (num2 == nums[i]) ++ time2;
        }
        if (time1 > nums.size() / 3) result.push_back(num1);
        if (time2 > nums.size() / 3) result.push_back(num2);
        return result;
    }
};
// Python代码
class Solution:
    # param {integer[]} nums
    # return integer[]
    def majorityElement(self, nums):
        if not nums:
            return []
        num1, num2, time1, time2 = None, None, 0, 0
        for num in nums:
            if num1 == num:
                time1 += 1
            elif num2 == num:
                time2 += 1
            elif time1 == 0:
                num1, time1 = num, 1
            elif time2 == 0:
                num2, time2 = num, 1
            else:
                time1, time2 = time1 - 1, time2 - 1
        numSize = len(nums)
        return [n for n in (num1, num2) if
                n is not None and nums.count(n) > numSize / 3]

四. 小结

关于摩尔投票法:http://mabusyao.iteye.com/blog/2223195

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