LeetCode 04 两个有序数组的中位数

ゝ一纸荒年。 2022-03-08 03:00 214阅读 0赞

**4. Median of Two Sorted Arrays    难度：Hard**

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]g
    nums2 = [3, 4]
    
    The median is (2 + 3)/2 = 2.5

翻译： 求两个排序数组的中位数

最容易想到的就是新建一个数组，将前两个数组的值挨个放进去，最后找到中位数。这样也能得到答案，但是时间复杂度是

O（m+n）, 达不到要求的  O(log (m+n)) 、居然也通过了。。。

package pers.leetcode;
    
    /**
     * 求两个数组的中位数
     *
     * @author admin
     * @date 2019/3/14 9:25
     */
    public class MedianSortedArrays {
        public static void main(String[] args) {
            int[] nums1 = {1, 3};
            int[] nums2 = {2};
            System.out.println("中位数 : " + findMedianSortedArrays(nums1, nums2));
        }
    
        public static double findMedianSortedArrays(int[] nums1, int[] nums2){
            double median;
            int lengthSum = nums1.length + nums2.length;
            int[] nums3 = new int[lengthSum];
            int i = 0, j = 0;
            while (i < nums1.length && j < nums2.length){
                if (nums1[i] <= nums2[j]){
                    nums3[i+j] = nums1[i++];
                }else {
                    nums3[i+j] = nums2[j++];
                }
            }
            if (i == nums1.length){
                for (int count = j; count < nums2.length; count++){
                    nums3[i+count] = nums2[count];
                }
            }
            if (j == nums2.length){
                for (int count = i; count < nums1.length; count++){
                    nums3[j+count] = nums1[count];
                }
            }
            if (lengthSum % 2 == 0){
                double sum = nums3[lengthSum/2 - 1] + nums3[lengthSum/2];
                median = sum/2;
            }else {
                median = nums3[lengthSum/2];
            }
            return median;
        }
    }

这里给出推荐的解法：

class Solution {
        public double findMedianSortedArrays(int[] A, int[] B) {
            int m = A.length;
            int n = B.length;
            if (m > n) { // to ensure m<=n
                int[] temp = A; A = B; B = temp;
                int tmp = m; m = n; n = tmp;
            }
            int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
            while (iMin <= iMax) {
                int i = (iMin + iMax) / 2;
                int j = halfLen - i;
                if (i < iMax && B[j-1] > A[i]){
                    iMin = i + 1; // i is too small
                }
                else if (i > iMin && A[i-1] > B[j]) {
                    iMax = i - 1; // i is too big
                }
                else { // i is perfect
                    int maxLeft = 0;
                    if (i == 0) { maxLeft = B[j-1]; }
                    else if (j == 0) { maxLeft = A[i-1]; }
                    else { maxLeft = Math.max(A[i-1], B[j-1]); }
                    if ( (m + n) % 2 == 1 ) { return maxLeft; }
    
                    int minRight = 0;
                    if (i == m) { minRight = B[j]; }
                    else if (j == n) { minRight = A[i]; }
                    else { minRight = Math.min(B[j], A[i]); }
    
                    return (maxLeft + minRight) / 2.0;
                }
            }
            return 0.0;
        }
    }