标签：
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.
Hint:
超过⌊ n/k ⌋ 最多有(k-1)个结果。k=3时最多2个结果。
因此设置两个candidate进行判断。
注意：留到最后的candidate不代表真正的结果。举例[3,2,3]，2是candidate但不是结果。
class Solution { public: vector<int> majorityElement(vector<int>& nums) { vector<int> ret; if(nums.empty()) return ret; int candidate1 = nums[0]; int count1 = 1; int candidate2 = nums[0]; int count2 = 0; for(int i = 1; i < nums.size(); i ++) { if(count1 != 0 && count2 != 0) { if(nums[i] == candidate1) count1 ++; else if(nums[i] == candidate2) count2 ++; else { count1 --; count2 --; } } else if(count1 != 0 && count2 == 0) { if(nums[i] == candidate1) count1 ++; else { candidate2 = nums[i]; count2 = 1; } } else if(count1 == 0 && count2 != 0) { if(nums[i] == candidate2) count2 ++; else { candidate1 = nums[i]; count1 = 1; } } else { candidate1 = nums[i]; count1 = 1; } } if(count1 != 0) { count1 = 0; for(int i = 0; i < nums.size(); i ++) { if(nums[i] == candidate1) count1 ++; } if(count1 > nums.size()/3) ret.push_back(candidate1); } if(count2 != 0) { count2 = 0; for(int i = 0; i < nums.size(); i ++) { if(nums[i] == candidate2) count2 ++; } if(count2 > nums.size()/3) ret.push_back(candidate2); } return ret; } };
【LeetCode】229. Majority Element II
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原文地址：http://www.cnblogs.com/ganganloveu/p/4630978.html
