Problem Statement
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3] Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2] Output: [1,2]
Problem link
Video Tutorial
You can find the detailed video tutorial here
Thought Process
It‘s a follow up question on Majority Element, given the hint, there should only be at most 2 elements that can satisfy the requirement (You cannot have 3 elements that all appear more than n/3 times). We can perform the voting algorithm on two potential candidates and later do an actual sum to eliminate the false positive (e.g., [1, 2, 1] would have both 1 and 2 as potential candidate, but 2 should not be included)
You can refer to Majority Element for a detailed explanation on the voting algorithm.
An extension could be changing n/3 to n/k. We just need extra arrays to store votes and candidates. It will take O(Nk) time complexity
Solutions
1 public List<Integer> majorityElement(int[] nums) {
2 List<Integer> res = new ArrayList<>();
3
4 if (nums == null || nums.length == 0) {
5 return res;
6 }
7
8 int candidate1 = 1;
9 int candidate2 = 1;
10 int vote1 = 0;
11 int vote2 = 0;
12
13 for (int i = 0; i < nums.length; i++) {
14 if (nums[i] == candidate1) {
15 vote1++;
16 } else if (nums[i] == candidate2) {
17 vote2++;
18 } else if (vote1 == 0) { // the order of this if/else actually matters, you cannot put the 0 check first since [8, 8, 7, 7, 7]
19 candidate1 = nums[i];
20 vote1++;
21 } else if (vote2 == 0) {
22 candidate2 = nums[i];
23 vote2++;
24 } else {
25 vote1--;
26 vote2--;
27 }
28 }
29
30 /* This is what you should do if we want to put vote1 == 0 check at the beginning
31 for (int i = 0; i < nums.length; i++) {
32 if (vote1 == 0 && candidate2 != nums[i]) {
33 candidate1 = nums[i];
34 } else if (vote2 == 0 && candidate1 != nums[i]) {
35 candidate2 = nums[i];
36 }
37 if (nums[i] == candidate1) {
38 vote1++;
39 } else if (nums[i] == candidate2) {
40 vote2++;
41 } else {
42 vote1--;
43 vote2--;
44 }
45 }
46 */
47
48 int count1 = 0;
49 int count2 = 0;
50
51 // need another pass to filter between the 2 candidates, e.g., [3, 2, 3], 2 is candidate but not more than 1/3
52 for (int num : nums) {
53 if (num == candidate1) {
54 count1++;
55 } else if (num == candidate2) {
56 count2++;
57 }
58 }
59
60 if (count1 > nums.length / 3) {
61 res.add(candidate1);
62 }
63 if (count2 > nums.length / 3) {
64 res.add(candidate2);
65 }
66
67 return res;
68 }
Voting implementation
Time Complexity: O(N) where N is the array size
Space Complexity: O(1) Constant space
References
原文地址：https://www.cnblogs.com/baozitraining/p/12059554.html