image.png
0. 链接
1. 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 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
2. 思路1：二维矩阵当成一维数组+二分查找
- 按照从左到右、从上到下的顺序，将二维矩阵看成一个包含m*n个元素的一维数组，然后运用二分查找即可
3. 代码
# coding:utf8
from typing import List
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
row = len(matrix)
if row == 0:
return False
col = len(matrix[0])
left = 0
right = row * col - 1
while left <= right:
mid = (left + right) >> 1
i = mid // col
j = mid % col
if matrix[i][j] < target:
left = mid + 1
elif matrix[i][j] > target:
right = mid - 1
else:
return True
return False
def print_matrix(matrix):
for each in matrix:
print(each)
print('=' * 50)
def my_test(solution, matrix, target):
print('input:')
print_matrix(matrix)
print('output:{}'.format(solution.searchMatrix(matrix, target)))
solution = Solution()
my_test(solution, [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
], 3)
my_test(solution, [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
], 13)
my_test(solution, [], 0)
输出结果
input:
[1, 3, 5, 7]
[10, 11, 16, 20]
[23, 30, 34, 50]
==================================================
output:True
input:
[1, 3, 5, 7]
[10, 11, 16, 20]
[23, 30, 34, 50]
==================================================
output:False
input:
==================================================
output:False
4. 结果
image.png
