Majority Number
原题链接：http://lintcode.com/en/problem/majority-number/#
Given an array of integers, the majority number is the number that occurs more than half of the size of the array. Find it.
Example
For [1, 1, 1, 1, 2, 2, 2], return 1
Challenge
O(n) time and O(1) space
SOLUTION 1:
http://www.geeksforgeeks.org/majority-element/
这里是一篇论文： http://www.cs.utexas.edu/~moore/best-ideas/mjrty/
这里用的算法是：MJRTY - A Fast Majority Vote Algorithm
1. 简单来讲，就是不断对某个议案投票，如果有人有别的议案，则将前面认为的议案的cnt减1，减到0换一个议案。
如果存在majority number，那么这个议案一定不会被减到0，最后会胜出。
2. 投票完成后，要对majority number进行检查，以排除不存在majority number的情况。如 1，2，3，4这样的数列，是没有majory number的。
很简单，统计一下结果议案的票数，没有过半就是没有majority number.
摘录一段解释：
METHOD 3 (Using Moore’s Voting Algorithm)
This is a two step process.
1. Get an element occurring most of the time in the array. This phase
will make sure that if there is a majority element then it will return
that only.
2. Check if the element obtained from above step is majority element.
1. Finding a Candidate:
The algorithm for first phase that works in O(n) is known as Moore’s
Voting Algorithm. Basic idea of the algorithm is if we cancel out each
occurrence of an element e with all the other elements that are
different from e then e will exist till end if it is a majority element.
findCandidate(a[], size)
1. Initialize index and count of majority element
maj_index = 0, count = 1
2. Loop for i = 1 to size – 1
(a)If a[maj_index] == a[i]
count++
(b)Else
count--;
(c)If count == 0
maj_index = i;
count = 1
3. Return a[maj_index]
Above algorithm loops through each element and maintains a count of a[maj_index], If next element is same then increments the count, if next element is not same then decrements the count, and if the count reaches 0 then changes the maj_index to the current element and sets count to 1.
First Phase algorithm gives us a candidate element. In second phase we
need to check if the candidate is really a majority element. Second
phase is simple and can be easily done in O(n). We just need to check if
count of the candidate element is greater than n/2.
Example:
A[] = 2, 2, 3, 5, 2, 2, 6
Initialize:
maj_index = 0, count = 1 –> candidate ‘2?
2, 2, 3, 5, 2, 2, 6
Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 1
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 5 => maj_index = 3, count = 1
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 2 => maj_index = 4
2, 2, 3, 5, 2, 2, 6
Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 1
Finally candidate for majority element is 2.
First step uses Moore’s Voting Algorithm to get a candidate for majority element.
2. Check if the element obtained in step 1 is majority
printMajority (a[], size)
1. Find the candidate for majority
2. If candidate is majority. i.e., appears more than n/2 times.
Print the candidate
3. Else
Print "NONE"
1 package Algorithms.lintcode.math;
2
3 import java.util.ArrayList;
4
5 public class MajorityNumber {
6 /**
7 * @param nums: a list of integers
8 * @return: find a majority number
9 */
10 public int majorityNumber(ArrayList<Integer> nums) {
11 // write your code
12 if (nums == null || nums.size() == 0) {
13 // No majority number.
14 return -1;
15 }
16
17 int candidate = nums.get(0);
18
19 // The phase 1: Voting.
20 int cnt = 1;
21 for (int i = 1; i < nums.size(); i++) {
22 if (nums.get(i) == candidate) {
23 cnt++;
24 } else {
25 cnt--;
26 if (cnt == 0) {
27 candidate = nums.get(i);
28 cnt = 1;
29 }
30 }
31 }
32
33 // The phase 2: Examing.
34 cnt = 0;
35 for (int i = 0; i < nums.size(); i++) {
36 if (nums.get(i) == candidate) {
37 cnt++;
38 }
39 }
40
41 // No majory number.
42 if (cnt <= nums.size() / 2) {
43 return -1;
44 }
45
46 return candidate;
47 }
48 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/lintcode/math/MajorityNumber.java