标签：题目 nes style public div appears sort empty map
题目：
Given an array of size n, find the majority element. The majority element is the element that appears more than ? n/2 ? times.
You may assume that the array is non-empty and the majority element always exist in the array.
题解：
Solution 1 ()
class Solution { public: int majorityElement(vector<int>& nums) { int n = nums.size(); int cnt = 0, majority = 0; for(auto n : nums) { if(majority == n) cnt++; else { cnt--; if(cnt <= 0) { cnt = 1; majority = n; } } } return majority; } };
Solution 2 ()
class Solution { public: int majorityElement(vector<int>& nums) { unordered_map<int, int> counts; int n = nums.size(); for (int i = 0; i < n; i++) if (++counts[nums[i]] > n / 2) return nums[i]; } };
Solution 3 ()
class Solution { public: int majorityElement(vector<int>& nums) { sort(nums.begin(), nums.end()); return nums[nums.size() / 2]; } };
Solution 4 ()
class Solution { public: int majorityElement(vector<int>& nums) { int ret = 0; for(int i=0; i<32; ++i) { int ones = 0, zeros = 0; for(int j=0; j<nums.size(); ++j) { if((nums[j] & (1<<i)) != 0) ++ones; else ++zeros; } if(ones > zeros) ret |= (1<<i); } return ret; } };
Solution 5 ()
class Solution { public: int majorityElement(vector<int> nums) { int n = nums.size(); if(n <= 1) return nums[0]; int m1 = majorityElement(vector<int> (nums.begin(), nums.begin() + n/2)); int m2 = majorityElement(vector<int> (nums.begin() + n/2, nums.end())); if(m1 == m2) return m1; int cnt = 0; for(auto num:nums) { if(num == m1) cnt++; if(cnt > n/2) return m1; } return m2; } };
【LeetCode】169. Majority Element
标签：题目 nes style public div appears sort empty map
原文地址：http://www.cnblogs.com/Atanisi/p/6803708.html
