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.

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 = 12.  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 = 13.  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 majority2.  If candidate is majority. i.e., appears more than n/2 times.       Print the candidate3.  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 code12         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 }

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/lintcode/math/MajorityNumber.java

Lintcode: Majority Number 解题报告