[Array]240. Search a 2D Matrix I

• 分类：Array
• 时间复杂度: O(m+n)

240. 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.

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.

代码：

非常直接的方法:

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        #边界条件
        if (matrix==None or len(matrix)==0 or len(matrix[0])==0):
            return False
        
        m=len(matrix)
        n=len(matrix[0])
        i=0
        j=n-1
        
        while(i<m and j>=0):
            if matrix[i][j]==target:
                return True
            elif matrix[i][j]>target:
                j-=1
            else:
                i+=1
                
        return False

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讨论：

1.这道题我看到的解法好像和分治无关，不知道为什么这个题要分到分治这个类别里？好像和DP的关系更大？

240做法