LeetCode - Median of Two Sorted Arrays

int m = nums1.size(), n = nums2.size();

将求two arrays的中位数转化为找第(m+n+1)/2个数和第(m+n+2)/2个数，中位数为这两个数的平均值。

解法一  Olog(m+n)

如何寻找two arrays的第K个值？
每次process k/2个数，如果nums1中的第k/2个数

Base Cases
k永远大于0，所以k=1的时候就是base case：if(k==1) return min(nums1[i],nums2[j]);
其中一条array已经遍历完：if(i==n) return nums2[j+k-1];
                                                if(j==m) return nums1[i+k-1];

易错点：不是每次都是process k/2个数，因为有array不够长了，所以我们用 int kk = min(min(k/2, n), m);

class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& nums2) {
        int n = nums1.size()+nums2.size();
        if(n==0) return 0;
        return (findK(nums1, nums2, 0, 0, (n+1)/2)+findK(nums1, nums2, 0, 0, (n+2)/2))*0.5;
    }
    int findK(vector& nums1, vector& nums2, int i, int j, int k){
        int n = nums1.size();
        int m = nums2.size();
        if(i==n) return nums2[j+k-1];
        if(j==m) return nums1[i+k-1];
        if(k==1) return min(nums1[i],nums2[j]);
        int kk = min(min(k/2, n), m);
        if(nums1[i+kk-1]

解法二

class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& 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;
    }
};

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        m, n = len(nums1), len(nums2)
        mid1 = (m+n+1)//2
        mid2 = (m+n+2)//2
        return (self.getKth(nums1, nums2, mid1)+self.getKth(nums1, nums2, mid2))/2
    
    # k one-indexed
    def getKth(self, a, b, k):
        if not a:
            return b[k-1]
        if not b:
            return a[k-1]
        ai, bi = (len(a)+1)//2, (len(b)+1)//2
        ae, be = a[ai-1], b[bi-1]
        
        if ai+bi<=k:
            if ae

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        m, n = len(nums1), len(nums2)
        if m>n: return self.findMedianSortedArrays(nums2, nums1)
        l, r = 0, m
        while l<=r:
            partition1 = l + (r-l)//2
            partition2 = (m+n+1)//2-partition1
            L1 = nums1[partition1-1] if partition1 != 0 else float('-inf')
            L2 = nums2[partition2-1] if partition2 != 0 else float('-inf')
            R1 = nums1[partition1] if partition1 != m else float('inf')
            R2 = nums2[partition2] if partition2 != n else float('inf')
            
            if L1<=R2 and L2<=R1:
                if (m+n)%2==0:
                    return (max(L1, L2)+min(R1, R2))/2
                else:
                    return max(L1, L2)
            elif L1>R2:
                r = partition1-1
            else:
                l = partition1+1