标签：more targe .com sort vector element 比较 ems and
原题链接：https://leetcode.com/problems/majority-element/description/
要求：
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.
想法：
1.比较普通的想法，有一个标记，初始值为1，对应一个计数初始值为1.
遍历数组，若数组值和标记值相等，计数加一；否则计数减一，当计数为0时，更改标记值为当前数组值。代码如下：
1 class Solution { 2 public: 3 int majorityElement(vector<int>& nums) { 4 int res = nums[0]; 5 int count = 1; 6 for (int i = 1;i < nums.size(); ++i) { 7 if (res == nums[i]) count++; 8 else { 9 count--; 10 if (count == 0) { 11 res = nums[i]; 12 count = 1; 13 } 14 } 15 } 16 return res; 17 } 18 };
2.首先将整个数组排序，由题意返回数组中间值即可。代码如下：
class Solution { public: int majorityElement(vector<int>& nums) { sort(nums.begin(), nums.end()); return nums[nums.size() / 2]; } };
第二个方法代码量相对较少，容易想到，不过我比较了一下运行时间，还是第一个方法的时间少。
【LeetCode】169. Majority Element
标签：more targe .com sort vector element 比较 ems and
原文地址：http://www.cnblogs.com/jialei-bupt/p/7259400.html
