Question
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.
Analysis
Method 1 HashMap:
use HashMap to record how many time each element appear
Method 2 Boyer-Moore majority vote Algorithm:
Solution
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public List<Integer> (int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int n : nums) {
if (!map.containsKey(n)) {
map.put(n, 1);
} else {
map.put(n, map.get(n) + 1);
}
}
List<Integer> res = new ArrayList<>();
for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
if (entry.getValue() > nums.length / 3) {
res. add(entry.getKey());
}
}
return res;
}
public List<Integer> (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;
}
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