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

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        a=[]
        for i in matrix:
            a+=i
        return target in a

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

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

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