leetCode169.MajorityElement数组
169. Majority Element
创新互联公司长期为1000多家客户提供的网站建设服务,团队从业经验10年,关注不同地域、不同群体,并针对不同对象提供差异化的产品和服务;打造开放共赢平台,与合作伙伴共同营造健康的互联网生态环境。为海晏企业提供专业的网站设计、网站建设,海晏网站改版等技术服务。拥有十载丰富建站经验和众多成功案例,为您定制开发。
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.
思路1:
使用map来处理。
class Solution { public: int majorityElement(vector& nums) { map record; for (int i = 0; i < nums.size(); i++) { if (record.find(nums[i]) == record.end()) { record.insert(pair (nums[i], 1)); if (record[nums[i]] > nums.size() / 2) { return nums[i]; } } else { record[nums[i]] += 1; if (record[nums[i]] > nums.size() / 2) { return nums[i]; } } } return 0; } };
思路2:
采用双循环
int majorityElement(vector& nums) { int hit = 0; int currentElem; for (int i = 0; i < nums.size(); i++) { currentElem = nums[i]; hit = 0; for (int j = 0; j < nums.size(); j++) { if (nums[j] == currentElem) { hit++; if (hit > nums.size() / 2) { return currentElem; } } } } return 0; }
2016-08-10 12:19:37
分享名称:leetCode169.MajorityElement数组
文章网址:http://cdiso.cn/article/jicoce.html