leetcode 日经贴，Cpp code -Majority Element II-优快云博客

本文介绍了一种寻找数组中出现次数超过元素总数三分之一的多数元素的算法实现。通过两轮遍历的方式，首先缩小候选范围，再确认最终结果。此方法适用于C++环境。

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class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        int n = nums.size();
        if (n <= 1) {
            return nums;
        }
        vector<int> ret;
        vector<pair<int, int> > candidates;
        candidates.resize(2);
        for (int i = 0; i < n; ++i) {
            int j = 0;
            for (; j < 2; ++j) {
                if (candidates[j].first && candidates[j].second == nums[i]) {
                    ++candidates[j].first;
                    break;
                }
            }
            if (j == 2) {
                for (j = 0; j < 2; ++j) {
                    if (candidates[j].first == 0) {
                        candidates[j].first = 1;
                        candidates[j].second = nums[i];
                        break;
                    }
                }
            }
            if (j == 2) {
                for (j = 0; j < 2; ++j) {
                    --candidates[j].first;
                }
            }
        }
        sort(candidates.begin(), candidates.end(), greater<pair<int,int> >());
        if (!candidates[0].first) {
            return ret;
        } else if (!candidates[1].first) {
            candidates.resize(1);
        }
        for (int i = 0; i < candidates.size(); ++i) {
            candidates[i].first = 0;
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < candidates.size(); ++j) {
                if (nums[i] == candidates[j].second) {
                    ++candidates[j].first;
                }
            }
        }
        for (int i = 0; i < candidates.size(); ++i) {
            if (candidates[i].first > n / 3) {
                ret.push_back(candidates[i].second);
            }
        }
        return ret;
    }
};