[leetcode] 74. Search a 2D Matrix @ python

原题

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 from left to right.
The first integer of each row is greater than the last integer of the previous row.
Example 1:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
Example 2:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false

解法1

先确定target所在的行, 即它<=该行最右边的数. 找到行之后进行二分查找.
Time: O(m + log(n))
Space: O(1)
代码

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        # base case
        if not matrix: return False
        row, col = len(matrix), len(matrix[0])
        # edge case
        if row == 0 or col == 0: 
            return False
        r, c = 0, col-1
        for i in range(row):
            if target <= matrix[i][c]:
                # binary search
                l, r = 0, col-1
                while l <= r:
                    mid = (l+r)//2
                    if matrix[i][mid] == target:
                        return True
                    if target < matrix[i][mid]:
                        r = mid-1
                    else:
                        l = mid+1
                return False
            else:
                continue
                
        return False

解法2

逐行查找
Time: O(m*n)
Space: O(1)

代码

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        # base case
        if not matrix: return False        
        for row in matrix:
            if target in row:
                return True
        return False