 6/21/2017 - 1:48 PM

## 240. Search a 2D Matrix II

1. Search a 2D Matrix II
``````class Solution(object):
def searchMatrix(self, mat, target):
"""
:type mat: List[List[int]]
:type target: int
:rtype: bool
"""
if len(mat) == 0 or len(mat) == 0:
return False
rowNum = len(mat)
colNum = len(mat)
rowIndex = 0
colIndex = colNum - 1
while colIndex >= 0 and rowIndex <= rowNum - 1:
# Like previous `Search a 2D Matrix` problem,
# except that the while loop will go for many times rather than just 1 or 2 times.
if mat[rowIndex][colIndex] < target:
rowIndex += 1
elif mat[rowIndex][colIndex] > target:
colIndex -= 1
else:
return True
return False``````

https://leetcode.com/problems/search-a-2d-matrix-ii/#/description

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.