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p0229_majority_element_ii.rs
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/**
* [229] Majority Element II
*
* Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
* Follow-up: Could you solve the problem in linear time and in O(1) space?
*
* Example 1:
*
* Input: nums = [3,2,3]
* Output: [3]
*
* Example 2:
*
* Input: nums = [1]
* Output: [1]
*
* Example 3:
*
* Input: nums = [1,2]
* Output: [1,2]
*
*
* Constraints:
*
* 1 <= nums.length <= 5 * 10^4
* -10^9 <= nums[i] <= 10^9
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/majority-element-ii/
// discuss: https://leetcode.com/problems/majority-element-ii/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn majority_element(nums: Vec<i32>) -> Vec<i32> {
let mut candidate1 = None;
let mut candidate2 = None;
let mut count1 = 0usize;
let mut count2 = 0usize;
for &num in &nums {
if Some(num) == candidate1 {
count1+=1;
} else if Some(num) == candidate2 {
count2+=1;
} else if count1 == 0 {
candidate1 = Some(num);
count1=1;
} else if count2 == 0 {
candidate2 = Some(num);
count2=1;
} else {
count1-=1;
count2-=1;
}
// if count1 == 0 {candidate1=None;}
// if count2 == 0 {candidate2=None;}
}
let mut result = vec![];
if let Some(candidate_num) = candidate1 {
if nums.iter().filter(|&x|{*x==candidate_num}).count() > nums.len()/3 {
result.push(candidate_num);
}
}
if let Some(candidate_num) = candidate2 {
if nums.iter().filter(|&x|{*x==candidate_num}).count() > nums.len()/3 {
result.push(candidate_num);
}
}
result
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_229() {
// assert_eq!(
// Solution::majority_element(vec![1, 1, 1, 2, 2, 2, 3, 3, 3]),
// vec![]
// );
assert_eq!(
Solution::majority_element(vec![1, 1, 1, 2, 2, 3, 3, 3]),
vec![1, 3]
);
assert_eq!(Solution::majority_element(vec![1]), vec![1]);
assert_eq!(Solution::majority_element(vec![5, 6, 6]), vec![6]);
// assert_eq!(Solution::majority_element(vec![1, 2, 3, 4]), vec![]);
}
}