LeetCode每日一题(Find Median from Data Stream)

最新推荐文章于 2025-02-03 15:37:10 发布

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该博客介绍了如何实现一个名为MedianFinder的类，用于处理数据流并在任何时候找到当前所有元素的中位数。通过二分插入排序的方法保持数据的有序性，从而快速找到中位数。在数据结构中，当元素数量为偶数时，中位数是中间两个数的平均值；当元素数量为奇数时，中位数是中间的那个数。题目还提出了针对数据范围优化解决方案的思考题。

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value and the median is the mean of the two middle values.

For example, for arr = [2,3,4], the median is 3.
For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5.

Implement the MedianFinder class:

MedianFinder() initializes the MedianFinder object.
void addNum(int num) adds the integer num from the data stream to the data structure.
double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input
["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"]
[[], [1], [2], [], [3], []]
Output
[null, null, null, 1.5, null, 2.0]

Explanation
MedianFinder medianFinder = new MedianFinder();
medianFinder.addNum(1);    // arr = [1]
medianFinder.addNum(2);    // arr = [1, 2]
medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)
medianFinder.addNum(3);    // arr[1, 2, 3]
medianFinder.findMedian(); // return 2.0

Constraints:

-105 <= num <= 105
There will be at least one element in the data structure before calling findMedian.
At most 5 * 104 calls will be made to addNum and findMedian.

Follow up:

If all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?
If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?

这题你要是不考虑后面的附加内容，那它就可以简化成一个维护有序数组的问题。二分法找到合适的位置插入，插入的时候注意插入的位置是在目标元素之前还是在目标元素之后就可以了。

struct MedianFinder {
    list: Vec<i32>,
}

impl MedianFinder {
    fn new() -> Self {
        Self { list: Vec::new() }
    }

    fn binary_insert(&mut self, start: usize, end: usize, num: i32) {
        if start == end {
            if self.list[start] > num {
                self.list.insert(start, num);
            } else {
                self.list.insert(start + 1, num);
            }
            return;
        }
        let mid = start + (end - start) / 2;
        if self.list[mid] < num {
            self.binary_insert(mid + 1, end, num);
        } else if self.list[mid] > num {
            self.binary_insert(start, mid, num);
        } else {
            self.list.insert(mid + 1, num);
        }
    }

    fn add_num(&mut self, num: i32) {
        if self.list.is_empty() {
            self.list.push(num);
            return;
        }
        self.binary_insert(0, self.list.len() - 1, num)
    }

    fn find_median(&self) -> f64 {
        if self.list.is_empty() {
            return 0.0;
        }
        let mid_idx = (self.list.len() - 1) / 2;
        if self.list.len() % 2 == 0 {
            return (self.list[mid_idx] + self.list[mid_idx + 1]) as f64 / 2.0;
        } else {
            return self.list[mid_idx] as f64;
        }
    }
}