• [x] Important!!!
• [x] 两种方法 都要掌握📈

There are two sorted arraysnums1andnums2of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

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

time : O(log min(m, n))

space O(1)

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if (nums1.length > nums2.length) {
            return findMedianSortedArrays(nums2, nums1);
        }
        int cut1 = 0;
        int cut2 = 0;
        int lbound = 0;
        int rbound = nums1.length;
        int len = nums1.length + nums2.length;
        while (cut1 <= nums1.length) {
            cut1 = lbound + (rbound - lbound) / 2;
            cut2 = len / 2 - cut1;
            double L1 = (cut1 == 0) ? Integer.MIN_VALUE : nums1[cut1 - 1];
            double L2 = (cut2 == 0) ? Integer.MIN_VALUE : nums2[cut2 - 1];
            double R1 = (cut1 == nums1.length) ? Integer.MAX_VALUE : nums1[cut1];
            double R2 = (cut2 == nums2.length) ? Integer.MAX_VALUE : nums2[cut2];
            if (L1 > R2) {
                rbound = cut1 - 1;
            }else if (L2 > R1) {
                lbound = cut1 + 1;
            }else{
                if (len % 2 == 0) {
                    L1 = L1 < L2 ? L2 : L1;
                    R1 = R1 < R2 ? R1 : R2;
                    return (L1 + R1) / 2.0;
                }else{
                    R1 = R1 < R2 ? R1 : R2;
                    return R1;
                }
            }

        } 
        return -1;
    }
}

对于一个长度为n的已排序数列a，若n为奇数，中位数为a[n / 2 + 1], 若n为偶数，则中位数(a[n / 2] + a[n / 2 + 1]) / 2; 如果我们可以在两个数列中求出第K小的元素，便可以解决该问题; 不妨设数列A元素个数为n，数列B元素个数为m，各自升序排序，求第k小元素; 取A[k / 2] B[k / 2] 比较; 如果 A[k / 2] > B[k / 2] 那么，所求的元素必然不在B的前k / 2个元素中(证明反证法); 反之，必然不在A的前k / 2个元素中，于是我们可以将A或B数列的前k / 2元素删去，求剩下两个数列的; k - k / 2小元素，于是得到了数据规模变小的同类问题，递归解决; 如果 k / 2 大于某数列个数，所求元素必然不在另一数列的前k / 2个元素中，同上操作就好。

class Solution {
    /**
     * @param A: An integer array.
     * @param B: An integer array.
     * @return: a double whose format is *.5 or *.0
     */
    public double findMedianSortedArrays(int[] A, int[] B) {
        int totalLength = A.length + B.length;
        if (totalLength % 2 == 1) {
            return findKth(A, 0, B, 0, totalLength / 2 + 1);
        }
        return (findKth(A, 0, B, 0, totalLength / 2) + findKth(A, 0, B, 0, totalLength / 2 + 1)) / 2.0;
    }

    public int findKth(int[] A, int A_start, int[] B, int B_start, int k) {
        if (A_start >= A.length) {
            return B[B_start + k - 1];
        }
        if (B_start >= B.length) {
            return A[A_start + k - 1];
        }

        if (k == 1) {
            return Math.min(A[A_start], B[B_start]);
        }


        int A_key = A_start + k / 2 - 1 < A.length
                ? A[A_start + k / 2 - 1]
                : Integer.MAX_VALUE;
        int B_key = B_start + k / 2 - 1 < B.length
                ? B[B_start + k / 2 - 1]
                : Integer.MAX_VALUE;

        if (A_key < B_key) {
            return findKth(A, A_start + k / 2, B, B_start, k - k / 2);
        } else {
            return findKth(A, A_start, B, B_start + k / 2, k - k / 2);
        }

    }
}