[LeetCode] 4. Median of Two Sorted Arrays 两个有序数组的中位数

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

求两个有序数组的中位数，限制了时间复杂度O(log (m+n))。

如果m+n为奇数，则中位数为AB[(m+n)/2 + 1]，偶数的话，中位数为(AB[(m+n)/2] + AB[(m+n)/2 + 1])/2。

如果没有时间复杂度O(log(m+n))的要求，就可以对两个数组使用归并排序，再找到他们的中位数，直接遍历两个数组查找，用2个变量分别指向两个数组，每次取较小的一个，然后将其指针后移动，直到找到中位数，时间复杂度为O(m+n)。但不是题目想要考察的。

解法：二分搜索Binary Search，T：O(log(m+n))。两个有序数组A(m), B(n)，k = (m+n)/2，奇数时找k+1大的数，偶数是找第k大和第k+1大的数在除2。找第k((m+n)/2)大的数。先在A，B中分别找第k/2大的数，如果A[k/2-1]==B[k/2-1]，那么这个数就是两个数组中第k大的数。如果A[k/2-1]<B[k/2-1], 那么说明A[0]到A[k/2-1]都不可能是第k大的数，所以需要舍弃这k/2，继续从A[k/2]到A[A.length-1]继续找。当然，因为这里舍弃了A[0]到A[k/2-1]这k/2个数，那么第k大也就变成了第k-k/2个大的数了。如果 A[k/2-1]>B[k/2-1]，那么说明B[0]到B[k/2-1]都不可能是第k大的数，舍弃这k/2。如此迭代或者递归操作，如果有一个数组为空了，则返回另一个数组的第k大(剩下需要二分长度)的数。如果k==1，只需返回此时所以数中排第一小的数，就返回此时A，B中第一个元素小的那个。

参考：爱做饭的小莹子

Java:
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public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
        int len = A.length + B.length;
        if (len % 2 == 1) {
            return findKth(A, 0, B, 0, len / 2 + 1);
        }
        return (
            findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)
        ) / 2.0;
    }
 
    public static 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);
        }
    }
}　　

Python:

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class Solution:
    # @return a float
    # @line20 must multiply 0.5 for return a float else it will return an int
    def getKth(self, A, B, k):
        lenA = len(A); lenB = len(B)
        if lenA > lenB: return self.getKth(B, A, k)
        if lenA == 0: return B[k - 1]
        if k == 1: return min(A[0], B[0])
        pa = min(k/2, lenA); pb = k - pa
        if A[pa - 1] <= B[pb - 1]:
            return self.getKth(A[pa:], B, pb)
        else:
            return self.getKth(A, B[pb:], pa)
     
    def findMedianSortedArrays(self, A, B):
        lenA = len(A); lenB = len(B)
        if (lenA + lenB) % 2 == 1:
            return self.getKth(A, B, (lenA + lenB)/2 + 1)
        else:
            return (self.getKth(A, B, (lenA + lenB)/2) + self.getKth(A, B, (lenA + lenB)/2 + 1)) * 0.5 

Python:

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class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        len1, len2 = len(nums1), len(nums2)
        if (len1 + len2) % 2 == 1:
            return self.getKth(nums1, nums2, (len1 + len2)/2 + 1)
        else:
            return (self.getKth(nums1, nums2, (len1 + len2)/2) + \
                    self.getKth(nums1, nums2, (len1 + len2)/2 + 1)) * 0.5
 
    def getKth(self, A, B, k):
        m, n = len(A), len(B)
        if m > n:
            return self.getKth(B, A, k)
 
        left, right = 0, m   
        while left < right:
            mid = left + (right - left) / 2
            if 0 <= k - 1 - mid < n and A[mid] >= B[k - 1 - mid]:
                right = mid
            else:
                left = mid + 1
 
        Ai_minus_1 = A[left - 1] if left - 1 >= 0 else float("-inf")
        Bj = B[k - 1 - left] if k - 1 - left >= 0 else float("-inf")
 
        return max(Ai_minus_1, Bj)

C++:

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class Solution {
public:
    double findKth(vector<int>& A, vector<int>& B, int A_st, int B_st, int k) {
        // 边界情况，任一数列为空
        if (A_st >= A.size()) {
            return B[B_st + k - 1];
        }
        if (B_st >= B.size()) {
            return A[A_st + k - 1];
        }
        // k等于1时表示取最小值，直接返回min
        if (k == 1) return min(A[A_st], B[B_st]);
        int A_key = A_st + k / 2 - 1 >= A.size() ? INT_MAX : A[A_st + k / 2 - 1];
        int B_key = B_st + k / 2 - 1 >= B.size() ? INT_MAX : B[B_st + k / 2 - 1];
        if (A_key < B_key){
            return findKth(A, B, A_st + k / 2, B_st, k - k / 2);
        } else {
            return findKth(A, B, A_st, B_st + k / 2, k - k / 2);
        }
         
    }
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int sum = nums1.size() + nums2.size();
        double ret;
         
        if (sum & 1) {
            ret = findKth(nums1, nums2, 0, 0, sum / 2 + 1);
        } else {
            ret = ((findKth(nums1, nums2, 0, 0, sum / 2)) +
                    findKth(nums1, nums2, 0, 0, sum / 2 + 1)) / 2.0;
        }
        return ret;
    }
};

　　　

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