解题思路:如果数组中确实有出现次数超一半的那个数，那数组排个序后，中间的那个就是了。
#include <iostream> #include<vector> #include<algorithm> using namespace std; class Solution { public: //排个序，如果存在出现次数超过一半的数，那么最中间呢个肯定就是了 int MoreThanHalfNum_Solution(vector<int> numbers) { if (numbers.empty()) { return 0; } sort(numbers.begin(), numbers.end()); // 排序，取数组中间那个数 int middle = numbers[numbers.size() / 2]; int count = 0; // 出现次数 for (size_t i = 0; i < numbers.size(); ++i) { if (numbers[i] == middle) ++count; } if (count >= (int)numbers.size() / 2) { return middle; } return 0; } }; int main() { vector<int> arr; arr.push_back(1); arr.push_back(2); arr.push_back(9); arr.push_back(9); arr.push_back(2); arr.push_back(9); Solution v; int u = v.MoreThanHalfNum_Solution(arr); printf("%d", u); return 0; }