【LeetCode】169.Majority Element (Easy)
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题目要求:n个元素的数组,找出现次数大于[n/2]的数(保证存在)
用map可以过:
int majorityElement(vector<int>& nums) { map<int, int> m; for (int i = 0; i != nums.size(); i++) m[nums[i]]++; for (map<int, int>::iterator i = m.begin(); i != m.end(); i++) if (i->second > nums.size() / 2) return i-> first; return -1;
Discuss里看到的做法:
int major=num[0], count = 1;for(int i=1; i<num.length;i++){ if(count==0) { count++; major=num[i]; } else if(major==num[i]) count++; else count--;}return major;相邻两个元素不一样就抵消掉,majority一定出现次数大于剩下才元素出现次数之和,不会都被抵消,剩下的就是majority。
只在majority存在的时候可以用。{3, 3, 4, 5, 6},3出现最多,但不是majority,返回6
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