Leetcode-240题:Search a 2D Matrix

题目

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.

思路

从矩阵右上角开始考虑，如果当前元素比所求大，跳转到左一列；如果比所求小，跳到下一行；如果是所求，直接返回。

代码

class Solution(object):

    def binarySearch(self, nums, l, r, target):
        if l > r:
            return False
        m = l + (r-l)/2
        if nums[m] == target:
            return True
        elif nums[m] > target:
            return self.binarySearch(nums, l, m-1, target)
        else:
            return self.binarySearch(nums, m+1, r, target)

    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if matrix==None or len(matrix)==0:
            return False
        m = len(matrix)
        n = len(matrix[0])
        i = 0
        j = n-1
        while i!=m-1 and j!=0:
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] < target:
                i += 1
            else:
                j -= 1
        if i == m-1:
            return self.binarySearch(matrix[i][:j+1],0,j,target)
        else:
            return self.binarySearch([matrix[k][j] for k in range(i,m)],0,m-i-1,target)