229. Majority Element II

发表于 | 分类于

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.

解法1：

Moore Voting 算法的延伸。 和Majority Element I类似，这里用两个变量来维护当前出现次数较大的数字。
对于第二个元素n2初始化为0是没关系的，因为他的计数器初始化为0。如果有数字和他一样的话计数器再加1。
C++

1

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if (nums == null || nums.length == 0) {
            return res;
        }
        int n1 = nums[0], n2 = 0, c1 = 0, c2 = 0;
        for (int num : nums) {
            if (num == n1) {
                c1++;
            } else if (num == n2) {
                c2++;
            } else if (c1 == 0) {
                n1 = num;
                c1 = 1;
            } else if (c2 == 0) {
                n2 = num;
                c2 = 1;
            } else {
                c1--;
                c2--;
            }
        }
        c1 = 0;
        c2 = 0;
        for (int num : nums) {
            if (num == n1) {
                c1++;
            } else if (num == n2) {
                c2++;
            }
        }
        if (c1 > nums.length / 3) {
            res.add(n1);
        }
        if (c2 > nums.length / 3) {
            res.add(n2);
        }
        return res;
    }
}