4. Median of Two Sorted Arrays

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;
    }
}