Majority Voting Algorithm也叫作Moore Voting Algorithm
在一个数组中,元素个数为n(假设最多投票元素存在),输出元素出现次数大于n/2的数
算法思路:1、一个变量cand表示所求的元素,一个变量count统计个数,将count初始化为0.
2、在遍历数组的过程上
(1)如果count=0,则将count=1,cand=array[I];
(2)否则,如果array[I]=cand,将count++,否则count--
代码如下:
class Solution { public: int majorityElement(vector<int>& nums) { int cand = -1; int count = 0; int len = nums.size(); for (int i = 0; i < len; i++) { if (count == 0) { count = 1; cand = nums[i]; } else if (nums[i] == cand) count++; else count--; } return cand; } };
一种更通用的情况为:
要求输出出现次数为n/k的元素
思路:需要维持一个长度为k-1的候选者数组及统计数组。如果候选者数组没有满,将其加入,相应的统计数计为1,如果在候选数组中出现过,将其计数加1,如果没有出现,将所有的计数减1
代码如下:
class Solution { public: vector<int> majorityElement(vector<int>& nums) { if (nums.empty()) return{}; return __majorityElement(nums, 3); } private: vector<int> __majorityElement(vector<int>& nums, int k) { int cnt = k - 1; vector<int> candidates(cnt, 0); vector<int> count(cnt, 0); for (int num : nums) { bool found = false; for (int i = 0; i < cnt; i++) { if (!count[i] || num == candidates[i]) { count[i]++; candidates[i] = num; found = true; break; } } if (!found) { for (int i = 0; i < cnt; i++) { count[i]--; } } } for (int i = 0; i < cnt; i++) { count[i] = 0; } for (int num : nums) { for (int i = 0; i < cnt; i++) { if (num == candidates[i]) { count[i]++; break; } } } vector<int> ans; for (int i = 0; i < cnt; i++) { if (count[i] > nums.size() / k) ans.push_back(candidates[i]); } return ans; } };
测试代码如下:
#include <iostream> #include <vector> using namespace std; int main() { Solution solver; vector<int> nums(1, 1); int ans = solver.majorityElement(nums); cout << ans << endl; return 0; }