题目

median-of-two-sorted-arrays

算法

* 直接模拟

* 直接模拟优化

* 二分

代码

* 直接模拟

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int total = nums1.size() + nums2.size();
        if (total % 2 == 1) {
            return findKth(nums1, 0, nums2, 0, total / 2 + 1);
        } else {
            return (findKth(nums1, 0, nums2, 0, total / 2) + findKth(nums1, 0, nums2, 0, total / 2 + 1)) / 2;
        }
    }
    double findKth(vector<int> &nums1, int i, vector<int> &nums2, int j, int k) {
        if (nums1.size() - i > nums2.size() - j) return findKth(nums2, j, nums1, i, k);
        if (nums1.size() == i) return nums2[j + k - 1];
        if (k == 1) return min(nums1[i], nums2[j]);
        int pa = min(i + k / 2, int(nums1.size())), pb = j + k - pa + i;
        if (nums1[pa - 1] < nums2[pb - 1]) 
            return findKth(nums1, pa, nums2, j, k - pa + i);
        else if (nums1[pa - 1] > nums2[pb - 1]) 
            return findKth(nums1, i, nums2, pb, k - pb + j);
        else 
            return nums1[pa - 1];
    }
};

* 直接模拟优化

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        return (findKth(nums1, nums2, (m + n + 1) / 2) + findKth(nums1, nums2, (m + n + 2) / 2)) / 2.0;
    }
    int findKth(vector<int> nums1, vector<int> nums2, int k) {
        int m = nums1.size(), n = nums2.size();
        if (m > n) return findKth(nums2, nums1, k);
        if (m == 0) return nums2[k - 1];
        if (k == 1) return min(nums1[0], nums2[0]);
        int i = min(m, k / 2), j = min(n, k / 2);
        if (nums1[i - 1] > nums2[j - 1]) {
            return findKth(nums1, vector<int>(nums2.begin() + j, nums2.end()), k - j);
        } else {
            return findKth(vector<int>(nums1.begin() + i, nums1.end()), nums2, k - i);
        }
        return 0;
    }
};

* 二分

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        if (m < n) return findMedianSortedArrays(nums2, nums1);
        if (n == 0) return ((double)nums1[(m - 1) / 2] + (double)nums1[m / 2]) / 2.0;
        int left = 0, right = n * 2;
        while (left <= right) {
            int mid2 = (left + right) / 2;
            int mid1 = m + n - mid2;
            double L1 = mid1 == 0 ? INT_MIN : nums1[(mid1 - 1) / 2];
            double L2 = mid2 == 0 ? INT_MIN : nums2[(mid2 - 1) / 2];
            double R1 = mid1 == m * 2 ? INT_MAX : nums1[mid1 / 2];
            double R2 = mid2 == n * 2 ? INT_MAX : nums2[mid2 / 2];
            if (L1 > R2) left = mid2 + 1;
            else if (L2 > R1) right = mid2 - 1;
            else return (max(L1, L2) + min(R1, R2)) / 2;
        }
        return -1;
    }
};