LeetCode----Search a 2D Matrix II

Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[  [1,   4,  7, 11, 15],  [2,   5,  8, 12, 19],  [3,   6,  9, 16, 22],  [10, 13, 14, 17, 24],  [18, 21, 23, 26, 30]]

Given target = 5, return true.

Given target = 20, return false.

分析：

又是一道二分查找问题，对于矩阵的每一行进行二分查找。

代码：

class Solution(object):    def searchMatrix(self, matrix, target):        """        :type matrix: List[List[int]]        :type target: int        :rtype: bool        """        m_len = len(matrix)  # number of lines        if not m_len:            return False        n_len = len(matrix[0])  # number of colums        if not n_len:            return False        notfound = True        for i in range(m_len):            if matrix[i][0] <= target <= matrix[i][n_len - 1] and notfound:                notfound = self.notFoundTarget(0, n_len - 1, matrix[i], target)            if not notfound:                break        return not notfound    def notFoundTarget(self, left, right, nums, target):        while left <= right:            mid = (left + right) / 2            if nums[mid] == target:                return False            elif nums[mid] > target:                right = mid - 1            else:                left = mid + 1        return True