4.Median of Two Sorted Arrays

Tags: Hard Array

Link: https://leetcode.com/problems/median-of-two-sorted-arrays/

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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.

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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

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Answer:

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;
    }
};