[LeetCode] Majority Element II

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.

分析：和Majority Number类似的思路。但是要考虑不存在的情况。时间复杂度O(n),空间复杂度O(1)

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        vector<int> res;
        if (nums.empty()) return res;
        if (nums.size() == 1) {
            res.push_back(nums[0]);
            return res;
        }
        
        int candidate1 = 0;
        int candidate2 = 0;
        int times1 = 0;
        int times2 = 0;
        
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == candidate1) {
                times1++;
            } else if (nums[i] == candidate2) {
                times2++;
            } else if (times1 == 0) {
                times1 = 1;
                candidate1 = nums[i];
            } else if (times2 == 0) {
                times2 = 1;
                candidate2 = nums[i];
            } else {
                times1--;
                times2--;
            }
        }
        
        times1 = 0;
        times2 = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == candidate1) {
                times1++;
            } else if (nums[i] == candidate2) {
                times2++;
            }
        }
        
        if (times1 > nums.size() / 3)
            res.push_back(candidate1);
        if (times2 > nums.size() / 3)
            res.push_back(candidate2);
        
        return res;
    }
};