LeetCode 229. Majority Element II（众数II）

原题网址：https://leetcode.com/problems/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.

方法一：先排序，然后统计。

public class Solution {    public List<Integer> majorityElement(int[] nums) {        List<Integer> results = new ArrayList<>();        if (nums == null || nums.length == 0) return results;        Arrays.sort(nums);        int prev = nums[0];        int count = 1;        for(int i=1; i<=nums.length; i++) {            if (i<nums.length) {                if (prev == nums[i]) count ++;                else {                    if (count > nums.length/3) results.add(prev);                    prev = nums[i];                    count = 1;                }            } else {                if (count > nums.length/3) results.add(prev);            }        }        return results;    }}

方法二：摩尔投票算法（Moore Vote Algorithm）

public class Solution {    public List<Integer> majorityElement(int[] nums) {        int num1 = 0, num2 = 1;        int count1 = 0, count2 = 0;        for(int num: nums) {            if (count1 == 0) {                num1 = num;                count1 = 1;            } else if (num1 == num) {                count1 ++;            } else if (count2 == 0) {                num2 = num;                count2 = 1;            } else if (num2 == num) {                count2 ++;            } else {                count1 --;                count2 --;                if (count1 == 0 && count2 > 0) {                    num1 = num2;                    count1 = count2;                    num2 = 0;                    count2 = 0;                }            }        }        if (count1 > 0) {            count1 = 0;            for(int num: nums) if (num1 == num) count1 ++;        }        if (count2 > 0) {            count2 = 0;            for(int num: nums) if (num2 == num) count2 ++;        }        List<Integer> results = new ArrayList<>();        if (count1*3>nums.length) results.add(num1);        if (count2*3>nums.length) results.add(num2);        return results;    }}

感谢网友，参考文章：

http://www.programcreek.com/2014/07/leetcode-majority-element-ii-java/

http://www.cnblogs.com/grandyang/p/4606822.html

http://bookshadow.com/weblog/2015/06/29/leetcode-majority-element-ii/

http://blog.csdn.net/xudli/article/details/46784149

http://codechen.blogspot.com/2015/06/leetcode-majority-element-ii.html

http://www.cs.utexas.edu/~moore/best-ideas/mjrty/

https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority_vote_algorithm

方法三：保存三个最高频数字作为候选，该方法与摩尔算法等效。

public class Solution {    public List<Integer> majorityElement(int[] nums) {        int[] counts = new int[3];        int[] recents = new int[3];        int size = 0;        for(int i=0; i<nums.length; i++) {            if (size > 0 && recents[0] == nums[i]) counts[0] ++;            else if (size > 1 && recents[1] == nums[i]) counts[1] ++;            else if (size > 2 && recents[2] == nums[i]) counts[2] ++;            else if (size < 3) {                recents[size] = nums[i];                counts[size] = 1;                size ++;            } else {                int min = 0;                if (counts[1] < counts[min]) min = 1;                if (counts[2] < counts[min]) min = 2;                recents[min] = nums[i];                counts[min] = 1;            }        }        List<Integer> results = new ArrayList<>();        for(int i=0; i<size; i++) {            counts[i] = 0;            for(int num: nums) if (recents[i] == num) counts[i] ++;            if (counts[i]>nums.length/3) results.add(recents[i]);        }        return results;    }}