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 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
Solutions
1. Binary Search
要找到trick,每一行最右边的数都比该行前面的数大,那么先可以通过每一行最后一个数确定行,然后采用 binary search 即可:
# Time: O(nlogm)
# Space: O(n)
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
if not matrix or len(matrix) == 0:
return False
h, w = len(matrix), len(matrix[0])
r = -1
for i in range(h):
if w-1 >= 0 and matrix[i][w-1] >= target:
r = i
break
if r == -1:
return False
if self.binary_search(matrix[r], target) == -1:
return False
else:
return True
def binary_search(self, nums, target):
if not nums:
return -1
n = len(nums)
l, r = 0, n - 1
while l <= r:
mid = (l + r) >> 1
if nums[mid] == target:
return mid
elif nums[mid] > target:
r = mid - 1
else:
l = mid + 1
return -1
# 136/136 cases passed (68 ms)
# Your runtime beats 92.86 % of python3 submissions
# Your memory usage beats 5.88 % of python3 submissions (14.9 MB)