lintcode:Majority Number II_appear k times lintcode-优快云博客

本文介绍了一种在O(n)时间内找到数组中出现次数超过1/3的多数元素的方法，并提供了一个具体的C++实现示例。

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

Find it.

Have you met this question in a real interview?

Yes

Notice

There is only one majority number in the array.

Example

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

Challenge

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

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: The majority number occurs more than 1/3.
     */
    int majorityNumber(vector<int> nums) {
        // write your code here
        
        int cond1;
        int count1 = 0;
        int cond2;
        int count2 = 0;
        
        for (int i=0; i<nums.size(); i++)
        {
            if (count1 == 0)
            {
                cond1 = nums[i];
                count1 = 1;
            }
            else if (nums[i] != cond1 && count2 == 0) //这个判断很重要
            {
                cond2 = nums[i];
                count2 = 1;
            }
            else
            {
                if (nums[i] == cond1)
                {
                    count1++;
                }
                else if (nums[i] == cond2)
                {
                    count2++;
                }
                else
                {
                    count1--;
                    count2--;
                }
            }
        }
        
        int t1 = 0;
        int t2 = 0;
        for (int i=0; i<nums.size(); i++)
        {
            if (count1 > 0 && nums[i] == cond1)
                t1++;
            else if (count2 > 0 && nums[i] == cond2)
                t2++;
        }
        
        if (t1 > nums.size()/3)
            return cond1;
        if (t2 > nums.size()/3)
            return cond2;
    }
};