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这是个young‘s矩阵,最好的就是O(m+n)的算法了
class Solution: # @param {integer[][]} matrix # @param {integer} target # @return {boolean} def searchMatrix(self, matrix, target): m = len(martix) if m == 0: return False n = len(matrix[0]) i, j = m - 1, 0 while i >= 0 and j < n: if matrix[i][j] == target: return True elif matrix[i][j] < target: j += 1 else: i -= 1 return False
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原文地址:http://www.cnblogs.com/dapanshe/p/4672300.html