LeetCode169. 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. （Easy）

You may assume that the array is non-empty and the majority element always exist in the array.

分析：

抵消的思想，維護一個result和count,與result相同則count++，否則count--，到0時更新result的值，由於主元素個數較多，所有最後一定能留下。

程式碼：

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

LintCode47. Majority NumberII

Given an array of integers, the majority number is the number that occurs more than 1/3

Find it.

Notice : There is only one majority number in the array. 

分析：

同majority number1的思路相似，維護兩個result和count，相同則對相應count操作，不同則均減一；

注意最後剩下兩個元素，並不一定count值大的就一定是出現次數多的（可能另一個參與抵消過多），所以需要重新遍歷一遍，對這個兩個數比較出現次數大小。

程式碼：

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers
 5      * @return: The majority number occurs more than 1/3.
 6      */
 7     int majorityNumber(vector<int> nums) {
 8         // write your code here
 9         int candidate1 = 0, candidate2 = 0;
10         int count1 = 0, count2 = 0;
11         for (int i = 0; i < nums.size(); ++i) {
12             if (nums[i] == candidate1 && count1 != 0) {
13                 count1++;
14             } 
15             else if (nums[i] == candidate2 && count2 != 0) {
16                 count2++;
17             }
18             else if (count1 == 0) {
19                 candidate1 = nums[i];
20                 count1 = 1;
21             }
22             else if (count2 == 0) {
23                 candidate2 = nums[i];
24                 count2 = 1;
25             }
26             else {
27                 count1--;
28                 count2--;
29             }
30         }
31         count1 = 0;
32         count2 = 0;
33         for (int i = 0; i < nums.size(); ++i) {
34             if (nums[i] == candidate1) {
35                 count1++;
36             }
37             else if (nums[i] == candidate2) {
38                 count2++;
39             }
40         }
41         return count1 > count2 ? candidate1 : candidate2;
42     }
43 };

LeetCode229. Majority Element 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.（Medium）

分析：

這個與lintcode中的majority number2基本相似，只是要求找到所有的大於n / 3次的元素（至多也就是兩個）；

所以最後一步從比較兩個canditate的count大小，變成將這兩個count與 size() / 3比較。

程式碼：

 1 class Solution {
 2 public:
 3     vector<int> majorityElement(vector<int>& nums) {
 4         int candidate1 = 0, candidate2 = 0;
 5         int count1 = 0, count2 = 0;
 6         for (int i = 0; i < nums.size(); ++i) {
 7             if (nums[i] == candidate1 && count1 != 0) {
 8                 count1++;
 9             } 
10             else if (nums[i] == candidate2 && count2 != 0) {
11                 count2++;
12             }
13             else if (count1 == 0) {
14                 candidate1 = nums[i];
15                 count1 = 1;
16             }
17             else if (count2 == 0) {
18                 candidate2 = nums[i];
19                 count2 = 1;
20             }
21             else {
22                 count1--;
23                 count2--;
24             }
25         }
26         count1 = 0;
27         count2 = 0;
28         for (int i = 0; i < nums.size(); ++i) {
29             if (nums[i] == candidate1) {
30                 count1++;
31             }
32             else if (nums[i] == candidate2) {
33                 count2++;
34             }
35         }
36         vector<int> result;
37         if (count1 > nums.size() / 3) {
38             result.push_back(candidate1);
39         }
40         if (count2 > nums.size() / 3) {
41             result.push_back(candidate2);
42         }
43         return result;
44 
45     }
46 };

LintCode48. Majority Number III

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array. （Medium）

Find it.

Notice：There is only one majority number in the array.

分析：

從前一題的1/3變為1/k，道理還是一樣，不過這次需要用一個hashmap來維護出現的次數，注意unordered_map插入刪除相關操作的寫法即可。

尤其hashmap元素個數等於k需要刪除的時候，需要維護一個vector存key，如果用iterator邊走邊刪除可能出現位置變化。

程式碼：

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers
 5      * @param k: As described
 6      * @return: The majority number
 7      */
 8     int majorityNumber(vector<int> nums, int k) {
 9         // write your code here
10         unordered_map<int, int> hash;
11         for (int i = 0; i < nums.size(); ++i) {
12             if (hash.size() < k) {
13                 hash[nums[i]]++;
14             }
15             else {
16                 vector<int> eraseVec;
17                 for (auto itr = hash.begin(); itr != hash.end(); ++itr) {
18                     (itr -> second)--;
19                     if (itr -> second == 0) {
20                         eraseVec.push_back(itr -> first);
21                     }
22                 }
23                 for (int i = 0; i < eraseVec.size(); ++i) {
24                     hash.erase(eraseVec[i]);
25                 }
26                 hash[nums[i]]++;
27             }
28         }
29         for (auto& n : hash) {
30             n.second = 0;
31         }
32         for (int i = 0; i < nums.size(); ++i) {
33             if (hash.find(nums[i]) != hash.end()) {
34                 hash[nums[i]]++;
35                 if (hash[nums[i]] > nums.size() / k) {
36                     return nums[i];
37                 }
38             }
39         }
40         return -1;
41     }
42 };