标签：solution public empty logs and log 不同的 pre div
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.
Moore voting algorithm--每找出两个不同的element，就成对删除即count--，最终剩下的一定就是所求的。时间复杂度：O(n)
class Solution { public: int majorityElement(vector<int>& nums) { int count = 0; int preNum = nums[0]; for(int i = 1; i < nums.size(); i++){ if(nums[i]==preNum){ count++; } else{ if(count == 0){ preNum = nums[i]; } else{ count--; } } } return preNum; } };
标签：solution public empty logs and log 不同的 pre div
原文地址：http://www.cnblogs.com/qionglouyuyu/p/6682896.html
