# Majority Number III
### Source
- lintcode: [(48) Majority Number III](http://www.lintcode.com/en/problem/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.
Find it.
Example
Given [3,1,2,3,2,3,3,4,4,4] and k=3, return 3.
Note
There is only one majority number in the array.
Challenge
O(n) time and O(k) extra space
~~~
### 题解
[Majority Number II](http://algorithm.yuanbin.me/zh-cn/math_and_bit_manipulation/majority_number_ii.html) 的升级版，有了前两道题的铺垫，此题的思路已十分明了，对 K-1个数进行相互抵消，这里不太可能使用 key1, key2...等变量，用数组使用上不太方便，且增删效率不高，故使用哈希表较为合适，当哈希表的键值数等于 K 时即进行清理，当然更准备地来讲应该是等于 K-1时清理。故此题的逻辑即为：1. 更新哈希表，若遇哈希表 size == K 时则执行删除操作，最后遍历哈希表取真实计数器值，返回最大的 key.
### Java
~~~
public class Solution {
/**
* @param nums: A list of integers
* @param k: As described
* @return: The majority number
*/
public int majorityNumber(ArrayList<Integer> nums, int k) {
HashMap<Integer, Integer> hash = new HashMap<Integer, Integer>();
if (nums == null || nums.isEmpty()) return -1;
// update HashMap
for (int num : nums) {
if (!hash.containsKey(num)) {
hash.put(num, 1);
if (hash.size() >= k) {
removeZeroCount(hash);
}
} else {
hash.put(num, hash.get(num) + 1);
}
}
// reset
for (int key : hash.keySet()) {
hash.put(key, 0);
}
for (int key : nums) {
if (hash.containsKey(key)) {
hash.put(key, hash.get(key) + 1);
}
}
// find max
int maxKey = -1, maxCount = 0;
for (int key : hash.keySet()) {
if (hash.get(key) > maxCount) {
maxKey = key;
maxCount = hash.get(key);
}
}
return maxKey;
}
private void removeZeroCount(HashMap<Integer, Integer> hash) {
Set<Integer> keySet = hash.keySet();
for (int key : keySet) {
hash.put(key, hash.get(key) - 1);
}
/* solution 1 */
Iterator<Map.Entry<Integer, Integer>> it = hash.entrySet().iterator();
while (it.hasNext()) {
Map.Entry<Integer, Integer> entry = it.next();
if(entry.getValue() == 0) {
it.remove();
}
}
/* solution 2 */
// List<Integer> removeList = new ArrayList<>();
// for (int key : keySet) {
// hash.put(key, hash.get(key) - 1);
// if (hash.get(key) == 0) {
// removeList.add(key);
// }
// }
// for (Integer key : removeList) {
// hash.remove(key);
// }
/* solution3 lambda expression for Java8 */
}
}
~~~
### 源码分析
此题的思路不算很难，但是实现起来还是有点难度的，**Java 中删除哈希表时需要考虑线程安全。**
### 复杂度分析
时间复杂度 O(n)O(n)O(n), 使用了哈希表，空间复杂度 O(k)O(k)O(k).
### Reference
- [Majority Number III 参考程序 Java/C++/Python](http://www.jiuzhang.com/solutions/majority-number-iii/)
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume