Majority Element

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.

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int>& nums) {
 4         if(nums.empty()) return 0;
 5         if(nums.size() == 1) return nums[0];
 6         
 7         int current;
 8         int count = 0;
 9         for(int i = 0; i < nums.size(); i++){
10             if(count == 0){
11                 current = nums[i];
12                 count = 1;
13             } 
14             else{
15                 if(nums[i] == current) count++;
16                 else count--;
17             }
18         }
19         return current;
20     }
21 };

Solution1:

 1 int majorityElement(vector<int> &num) {
 2         int len = num.size();
 3         for(int i = 0; i <= len / 2; i++){
 4             for(int j = i + 1; j < len - i; j++){
 5                 if(num[j] < num[i]){
 6                     int temp = num[i]; 
 7                     num[i] = num[j];
 8                     num[j] = num[i];
 9                 }
10             }
11         }
12         return num[len/2];
13     }

结果超时..参照https://leetcode.com/discuss/19151/solution-computation-space-problem-can-extended-situation解题思路，可得代码如下

Solution2:

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int> &num) {
 4         int len = num.size();
 5         int count = 0;
 6         int count_number = 0;
 7         for(int i = 0; i < len; i++){
 8             if(count == 0){
 9                 count_number = num[i];
10                 count++;
11             }
12             else{
13                 if(count_number == num[i]) count++;
14                 else count--;
15             }
16         }
17         return count_number;
18     }
19 };