Home » Python » [leetcode ][4] Median of Two Sorted A

There, are, two, sorted, nums1, and, of, size, m, and, N, respectively., Find, the, of, the, two, sorted, arrays., nums2, The, overall, run, time, complexity, should, arrays, be, O (log (m+n)).
. Median
https://leetcode.com/problems/median-of-two-sorted-arrays/
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Ideas: using two points to find

``class Solution (object) : def, findMedianSortedArrays (self, nums1, nums2) : """: type, nums1:, List[int]: type, nums2:, List[int]: rtype: float"" ""M = len (nums1); n = len (nums2)if (m+n)%2 = = 1:return, self.binarySearch (nums1, m, nums2, N, (m+n) /2+1)else:return, float ((self.binarySearch, nums1, m, nums2, N, (M +, n) / 2) + self.binarySearch (nums1, m, nums2, N, 2 (M + n) / 2 + 1))def, binarySearch (self, nums1, m, nums2, N, K) : if m<=0:return nums2[k-1]if n<=0:return nums1[k-1]if k<=1:return min (nums1[0], nums2[0])if nums1[m/2] > nums2[n/2]:if (m/2 n/2 + +1 > = k:)return, self.binarySearch (nums1, m/2, nums2, N, K)else:return, self.binarySearch (nums1, m, nums2[n/2+1: n], n-n/2-1, K - n/2, -1)else:if (m/2 n/2 + +1 > = k:)return, self.binarySearch (nums1, m, nums2, n/2, K)else:return, self.binarySearch (nums1[m/2+1: m], m-m/2-1, nums2, N, k-m/2-1)``