229. 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.
Boyer-Moore Majority Vote algorithm
The essential concepts is you keep a counter for the majority number X. If you find a number Y that is not X, the current counter should deduce 1. The reason is that if there is 5 X and 4 Y, there would be one (5-4) more X than Y. This could be explained as "4 X being paired out by 4 Y".
And since the requirement is finding the majority for more than ceiling of [n/3], the answer would be less than or equal to two numbers.
So we can modify the algorithm to maintain two counters for two majorities.
public List<Integer> majorityElement(int[] nums) {
List<Integer> list = new ArrayList<>();
if (nums == null || nums.length == 0) {
return list;
}
if (nums.length == 1) {
list.add(nums[0]);
return list;
}
int ans1 = 0, ans2 = 0;
int count1 = 0, count2 = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != ans1 && nums[i] != ans2) {
if (count1 == 0) {
ans1 = nums[i];
count1++;
} else if (count2 == 0) {
ans2 = nums[i];
count2++;
} else {
count1--;
count2--;
}
} else if (nums[i] == ans1) {
count1++;
} else {
count2++;
}
}
count1 = 0;
count2 = 0;
for (int i : nums) {
if (i == ans1) {
count1++;
} else if (i == ans2) {
count2++;
}
}
if (count1 > nums.length / 3) {
list.add(ans1);
}
if (count2 > nums.length / 3) {
list.add(ans2);
}
return list;
}
多个候选人的投票,从0开始计票, 因为每一票都要计算count， follow up 延伸到找K个 这样的数:
A simple method is to pick all elements one by one. For every picked element, count its occurrences by traversing the array, if count becomes more than n/k, then print the element. Time Complexity of this method would be O(n2).
A better solution is to use sorting. First, sort all elements using a O(nLogn) algorithm. Once the array is sorted, we can find all required elements in a linear scan of array. So overall time complexity of this method is O(nLogn) + O(n) which is O(nLogn).
A simple method is to pick all elements one by one. For every picked element, count its occurrences by traversing the array, if count becomes more than n/k, then print the element. Time Complexity of this method would be O(n2).
A better solution is to use sorting. First, sort all elements using a O(nLogn) algorithm. Once the array is sorted, we can find all required elements in a linear scan of array. So overall time complexity of this method is O(nLogn) + O(n) which is O(nLogn).
