74. 搜索二维矩阵
编写一个高效的算法来判断 m x n
矩阵中,是否存在一个目标值。该矩阵具有如下特性:
- 每行中的整数从左到右按升序排列。
- 每行的第一个整数大于前一行的最后一个整数。
示例 1:
输入:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3 输出:true
示例 2:
输入:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13 输出:false
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-104 <= matrix[i][j], target <= 104
0
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: a=[] for i in matrix: a+=i return target in a
1
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: def binarySearch(arr, l, r, x): while l <= r: mid = l + (r - l) // 2; # Check if x is present at mid if arr[mid] == x: return mid # If x is greater, ignore left half elif arr[mid] < x: l = mid + 1 # If x is smaller, ignore right half else: r = mid - 1 return -1 a=[] for i in matrix: a+=i return binarySearch(a, 0, len(a)-1, target)!=-1
2
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: def binarySearch(a, x): i = bisect_left(a, x) if i != len(a) and a[i] == x: return i else: return -1 a=[] for i in matrix: a+=i return binarySearch(a, target)!=-1
3
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: for i in range(len(matrix)): if i<len(matrix)-1: if matrix[i][0]<=target and matrix[i+1][0]>target: for j in matrix[i]: if j==target: return True else: if matrix[i][0]<=target and matrix[i][-1]>=target: for j in matrix[i]: if j==target: return True return False