[LeetCode] Majority Element II

Posted 2020-11-14 immjc

Given an integer array of size n, find all elements that appear more than ? n/3 ? times.

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

Input: [3,2,3]
Output: [3]

Example 2:

Input: [1,1,1,3,3,2,2,2]
Output: [1,2]

n/3 times说明这样的数至多存在2个。

所以使用两个数n1,n2表示候选的数字，c1,c2表示这两个数出现的次数。

遍历数组，记当前数组为num，如果num与n1,n2其一相等，则对应的c1或c2加一。

如果num与n1和n2都不等，此时如果c1或c2其一为0，则令对应的候选数为num，并令其加一，

否则c1和c2各自减一，

最后再统计一次候选数字出现的次数是否大于n/3

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        vector<int> res;
        if (nums.empty())
            return res;
        int n1 = nums[0], n2 = 0;
        int c1 = 1, c2 = 0;
        for (int i = 1; i < nums.size(); ++i)
        {
            if (nums[i] == n1)
                c1++;
            else if (nums[i] == n2)
                c2++;
            else
            {
                if (c1 == 0)
                {
                    n1 = nums[i];
                    c1 = 1;
                }
                else if (c2 == 0)
                {
                    n2 = nums[i];
                    c2 = 1;
                }
                else
                {
                    c1--;
                    c2--;
                }
            }
        }
        c1 = 0, c2 = 0;
        for (auto& num : nums)
        {
            if (num == n1)
                c1++;
            else if (num == n2)
                c2++;
        }
        if (c1 > nums.size() / 3)
            res.push_back(n1);
        if (c2 > nums.size() / 3 && n1 != n2)
            res.push_back(n2);
        return res;
    }
};
// 12 ms