
Search a 2D Matrix
发布日期:2021-05-13 00:13:03
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本文共 1795 字,大约阅读时间需要 5 分钟。
?????????m x n?????????????????????????????????????????????????????????????????????????????????????O(log m + log n)???????
?????????????????????????????????????????????????????????????????????????????????????????????????????
??????
class Solution: def midsearch(self, lists, target): left = 0 right = len(lists) - 1 while left <= right: mid = (left + right) // 2 if lists[mid] < target: left = mid + 1 elif lists[mid] == target: return mid else: right = mid - 1 return -1
?????
class Solution: def midsearch_1(self, lists, target): left = 0 right = len(lists) - 1 while left <= right: mid = (left + right) // 2 if target < lists[mid][0]: right = mid - 1 elif target > lists[mid][0]: left = mid + 1 else: return mid return -1
???
class Solution: def searchMatrix(self, matrix, target): if not matrix: return False rows = [row[0] for row in matrix] j = self.midsearch_1(rows, target) if j == -1: if target > matrix[-1][-1]: return False else: row = matrix[-1] i = self.midsearch(row, target) return i != -1 else: row = matrix[j] i = self.midsearch(row, target) return i != -1
????
midsearch????????????????????????????????????????????-1?
midsearch_1???????????????????????????????????????????????????
searchMatrix??????????????????????midsearch_1???????????????????????????????????midsearch??????
????????????????????????????????O(log m + log n)????????
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