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 1) 第一种思路为 因为每行是sorted, 并且后面的row要比前面的要大, 所以我们可以将它看做是一维的sorted array 去处理, 然后去Binary Search, 只是需要注意的是num = matrix[mid//lrc[1]:列数][mid%lrc[0]：行数]. 2） 第二种思路为， 找到可能在某一行（last index <= target），然后在该行中做常规的binary search
Code 1)
class Solution:
def searchMatrix(self, matrix, target):
if not matrix or len(matrix[0]) == 0: return False
lrc = [len(matrix), len(matrix[0])]
l, r = 0, lrc[0]*lrc[1] -1
while l + 1 < r:
mid = l + (r - l)//2
num = matrix[mid//lrc[1]][mid%lrc[1]]
if num > target:
r = mid
elif num < target:
l = mid
else:
return True
if target in [matrix[l//lrc[1]][l%lrc[1]], matrix[r//lrc[1]][r%lrc[1]]]:
return True
return False
2)
class Solution:
def searchMatrix(self, matrix, target):
if not matrix or len(matrix[0]) == 0 or matrix[0][0] > target or matrix[-1][-1] < target:
return False
colNums = list(zip(*matrix))[0] # first column
# find the last index <= target
l, r = 0, len(colNums) -1
while l + 1 < r:
mid = l + (r - l)//2
if colNums[mid] > target:
r = mid
elif colNums[mid] < target:
l = mid
else:
return True
rowNums = matrix[r] if colNums[r] <= target else matrix[l]
l, r = 0, len(rowNums) - 1
while l + 1 < r:
mid = l + (r -l)//2
if rowNums[mid] > target:
r = mid
elif rowNums[mid] < target:
l = mid
else:
return True
if target in [rowNums[l], rowNums[r]]: return True
return False