python二分法及小tips
原题在https://leetcode-cn.com/problems/median-of-two-sorted-arrays/
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class Solution: def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: nums1.extend(nums2) nums1.sort() len1 = len(nums1) return (nums1[len1//2 -1]+nums1[len1//2 ])/ 2.0 if len1 % 2 == 0 else nums1[int(len1/2)]
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class Solution: def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: n1 = len(nums1) n2 = len(nums2) if n1 > n2 : return self.findMedianSortedArrays(nums2,nums1) k = (n1 + n2 +1 )//2 left = 0 right = n1 while left < right : m1 = left + (right - left)//2 m2 = k - m1 if nums1[m1] <nums2[m2-1] : left = m1 + 1 else: right = m1 m1 = left m2 = k -m1 c1 = max(nums1[m1 - 1] if m1 > 0 else float("-inf"), nums2[m2 - 1] if m2 > 0 else float("-inf")) if (n1 + n2) % 2 == 1: return c1 c2 = min(nums1[m1] if m1 < n1 else float("inf"), nums2[m2] if m2 < n2 else float("inf")) return (c1+c2)/2
tips:
nums1.extend(nums2)
,nums1合取nums2 的数成为一个数组nums1.sort()
排序(默认升序,降序为nums1.sort(reverse = True)
)- float("-inf"),float("inf") 表示正负无穷
这个Leetcode的题目算法很有意思!!!