用 Rust 刷 leetcode 第四題

There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.

Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

use std::cmp::max;
use std::cmp::min;
pub fn find_median(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
    let n = nums1.len();
    let m = nums2.len();

    let mut l1 = 0;
    let mut l2 = 0;
    let mut r1 = 0;
    let mut r2 = 0;
    let mut c1: usize = 0;
    let mut c2: usize = 0;
    let mut lo: usize = 0;
    let mut hi: usize = 2*n;

    while lo <= hi { 
        c1 = (lo + hi)/2;
        c2 = m + n - c1;
        l1 = if (c1 == 0) {
            std::i32::MIN
        } else {
            nums1[(c1-1)/2]
        };

        r1 = if c1 == 2*n {
            std::i32::MAX  
        } else {
            nums1[c1/2]  
        };

        l2 = if (c2 == 0) {
            std::i32::MIN
        } else {
            nums2[(c2-1)/2]
        };

        r2 = if c2 == 2*m {
            std::i32::MAX  
        } else {
            nums2[c2/2]  
        };

        if l1 > r2 {
            hi = c1 - 1;
        } else if l2 > r1 {
            lo = c1 + 1;
        } else {
            break;
        }
    }

    ((max(l1, l2) + min(r1, r2)) as f64) / 2.0
}

impl Solution {   
    pub fn find_median_sorted_arrays(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
        let n = nums1.len();
        let m = nums2.len();

        if n > m {
            find_median(nums2, nums1)
        } else {
            find_median(nums1, nums2)
        }
    }
}

本作品採用《CC 協議》，轉載必須註明作者和本文連結