LeetCode OJ:Majority Element II(主元素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.

求主元素,这次的是大于sz/3就算是主元素,可以分为两轮查找,第一轮先查找元素数目较多的两个元素(可能不属于主元素),第二次再遍历来查找上面两个元素是否符合条件,代码如下:

 1 class Solution {
 2 public:
 3     vector<int> majorityElement(vector<int>& nums) {
 4         int m1, m2;
 5         int count1, count2;
 6         vector<int> ret;
 7         int sz = nums.size();
 8         if(!sz) return ret;
 9         m1 = nums[0];
10         m2 = 0;
11         count1 = 1;
12         count2 = 0;
13         for(int i = 1; i < sz; ++i){
14             if(m1 == nums[i])
15                 count1++;
16             else if(m2 == nums[i])
17                 count2++;
18             else if(count1 == 0){
19                 count1++;
20                 m1 = nums[i];
21             }else if(count2 == 0){
22                 count2++;
23                 m2 = nums[i];
24             }else{
25                 count1--;
26                 count2--;
27             }
28         }
29         count1 = count2 = 0;
30         for(int i = 0; i < sz; ++i){
31             if(nums[i] == m1) ++count1;
32             if(nums[i] == m2) ++count2;
33         }
34         if(count1 > sz/3) ret.push_back(m1);
35         if(m1 != m2)
36             if(count2 > sz/3) ret.push_back(m2);
37         return ret;
38     }
39 };
原文地址:https://www.cnblogs.com/-wang-cheng/p/4922892.html