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
解题思路:
这个 matrix 的排序比较特别,是从左到右递增,从上到下递增 可以从第一行的最右边开始查找,如果目标值大于矩阵值可以排除左边的,如果目标值小于矩阵值可以排除下面的
class Solution(object):
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
r = len(matrix)
if r == 0:
return False
c = len(matrix[0])
if c == 0:
return False
x , y = 0 , c - 1
while x < r and y >= 0:
# print(r,c)
print(x,y)
print(matrix[x][y])
if matrix[x][y] == target:
return True
elif matrix[x][y] < target:
x += 1
else:
y -= 1
return False
请多多指教。
文章标题:240. Search a 2D Matrix II
本文作者:顺强
发布时间:2019-04-03, 23:59:00
原始链接:http://shunqiang.ml/leetcode-240-search-a-2D-m-2/版权声明: "署名-非商用-相同方式共享 4.0" 转载请保留原文链接及作者。