229. Majority Element II

题目：

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.

链接： http://leetcode.com/problems/majority-element-ii/

题解：

依然是Boyer-Morrer Voting Algorithm。这回我们要设置两个变量来进行voting。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<>();
        if(nums == null || nums.length == 0)
            return res;
        int candidate1 = 0, candidate2 = 0, count1 = 0, count2 = 0;
        
        for(int i : nums) {
            if(i == candidate1)
                count1++;
            else if(i == candidate2)
                count2++;
            else if(count1 == 0) {
                candidate1 = i;
                count1 = 1;
            } else if (count2 == 0) {
                candidate2 = i;
                count2 = 1;
            } else {
                count1--;
                count2--;
            }
        }
        
        count1 = 0;
        count2 = 0;
        
        for(int i : nums) {
            if(i == candidate1)
                count1++;
            else if(i == candidate2)
                count2++;
        }
        
        if(count1 > nums.length / 3)
            res.add(candidate1);
        if(count2 > nums.length /3)
            res.add(candidate2);
        
        return res;
    }
}

二刷:

方法同一刷一样。一定要注意遍历数组的时候先判断num是否和num1和num2相等，再考虑count1 == 0 和count2 == 0的情况。最后便利完毕以后还要再加一个验证步骤。

Java:

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

三刷:

跟二刷一样。要注意先判断的是num和candidates的关系，其次再判断count

Java:

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<>();
        if (nums == null || nums.length == 0) return res;
        int candidate1 = 0, candidate2 = 0, count1 = 0, count2 = 0;
        for (int num : nums) {
            if (num == candidate1) {
                count1++;
            } else if (num == candidate2) {
                count2++;
            } else if (count1 == 0) {
                candidate1 = num;
                count1++;
            } else if (count2 == 0) {
                candidate2 = num;
                count2++;
            } else {
                count1--;
                count2--;
            }
        }
        count1 = 0;
        count2 = 0;
        for (int num : nums) {
            if (num == candidate1) count1++;
            else if (num == candidate2) count2++;
        }
        if (count1 > nums.length / 3) res.add(candidate1);
        if (count2 > nums.length / 3) res.add(candidate2);
        return res;
    }
}

Reference:

https://leetcode.com/discuss/42768/o-n-time-o-1-space

https://leetcode.com/discuss/49596/clear-o-n-solution-in-python-no-data-structure-or-sort

https://leetcode.com/discuss/46560/boyer-moore-method-java-implementation

https://leetcode.com/discuss/56737/my-o-n-time-solution-20ms

https://leetcode.com/discuss/43580/my-c-solution

https://leetcode.com/discuss/48238/c%23-solution-with-o-n-time-and-o-1-space-via-improved-quick-sort

https://leetcode.com/discuss/42806/boyer-moore-majority-vote-algorithm-generalization

https://leetcode.com/discuss/43248/boyer-moore-majority-vote-algorithm-and-my-elaboration