 LeetCode 229 Majority Element II（主要元素II）（Array）（Boyer–Moore majority vote algorithm）

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LeetCode 229 Majority Element II（主要元素II）（Array）（Boyer–Moore majority vote algorithm）

nomasp 2016-08-29 21:40:40 浏览347

原文

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.

Hint:

How many majority elements could it possibly have?

分析

The Boyer-Moore Vote Algorithm solves the majority vote problem in linear time O(n)and logarithmic space O(logn). The majority vote problem is to determine in any given sequence of choices whether there is a choice with more occurrences than half of the total number of choices in the sequence and if so, to determine this choice. Note how this definition contrasts with the mode in which it is not simply the choice with the most occurrences, but the number of occurrences relative to the total length of the sequence.

Mathematically, given a finite sequence (length n) of numbers, the object is to find the majority number defined as the number that appears more than n/2 times.

import java.util.*;
public class MajorityVote {
public int majorityElement(int[] num) {
int n = num.length;
int candidate = num, counter = 0;
for (int i : num) {
if (counter == 0) {
candidate = i;
counter = 1;
} else if (candidate == i) {
counter++;
} else {
counter--;
}
}

counter = 0;
for (int i : num) {
if (i == candidate) counter++;
}
if (counter <= n / 2) return -1;
return candidate;
}
public static void main(String[] args) {
MajorityVote s = new MajorityVote();
System.out.format("%d\n", s.majorityElement(new int[] {1, 2, 3}));
System.out.format("%d\n", s.majorityElement(new int[] {2, 2, 3}));
}
}

代码

Java

public class Solution {
public List<Integer> majorityElement(int[] nums) {
List<Integer> marjority = new ArrayList<>();
int n = nums.length;
int candidate1 = 0, candidate2 = 0, counter1 = 0, counter2 = 0;
for (int i : nums) {
if (candidate1 == i) {
counter1++;
} else if (candidate2 == i) {
counter2++;
} else if (counter1 == 0) {
candidate1 = i;
counter1 = 1;
} else if (counter2 == 0) {
candidate2 = i;
counter2 = 1;
} else {
counter1--;
counter2--;
}
}
counter1 = 0;
counter2 = 0;
for (int i : nums) {
if (i == candidate1) counter1++;
else if (i == candidate2) counter2++;
}
if (counter1 > n/3) marjority.add(candidate1);
if (counter2 > n/3) marjority.add(candidate2);
return marjority;
}
}

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