leetcode-240-搜索二维矩阵②

题目描述：

 

 最佳方法：O(m+n） O（1）

class Solution:
    def searchMatrix(self, matrix, target):
        if not matrix :
            return False

        row = len(matrix)
        col = len(matrix[0])
        i = 0
        j = col-1
        while i<row and j>=0:
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] > target:
                j -= 1
            else:
                i += 1
        return False

方法二：O(nlogn）*

class Solution:
    def searchMatrix(self, matrix, target):
        # an empty matrix obviously does not contain `target`
        if not matrix:
            return False

        def search_rec(left, up, right, down):
            # this submatrix has no height or no width.
            if left > right or up > down:
                return False
            # `target` is already larger than the largest element or smaller
            # than the smallest element in this submatrix.
            elif target < matrix[up][left] or target > matrix[down][right]:
                return False

            mid = left + (right-left)//2

            # Locate `row` such that matrix[row-1][mid] < target < matrix[row][mid]
            row = up
            while row <= down and matrix[row][mid] <= target:
                if matrix[row][mid] == target:
                    return True
                row += 1
            
            return search_rec(left, row, mid-1, down) or search_rec(mid+1, up, right, row-1)

        return search_rec(0, 0, len(matrix[0])-1, len(matrix)-1)