Leetcode 240. Search a 2D Matrix II

https://leetcode.com/problems/search-a-2d-matrix-ii/

Medium

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

---

• 三种解法
  • 1. 从右上角开始找，i.e. 当前行最大，当前列最小的开始。因为横竖都是升序，所以如果target > current，必然不在当前行；如果target < current，必然不在当前列。
  • 2. Python any() function
    • https://docs.python.org/3/library/functions.html#any
  • 3. 对每一行进行二分查找。记得Python3要用整除//, i.e. middle = (left + right) // 2。
  • https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/66140/My-concise-O(m%2Bn)-Java-solution
  • https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/66163/C%2B%2B-two-solutions-(O(m%2Bn)-O(mlogn))

class Solution:
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or len(matrix) == 0 or len(matrix[0]) == 0:
            return False
        
        row, column = 0, len(matrix[0]) - 1
        
        while row < len(matrix) and column >= 0:
            if target == matrix[row][column]:
                return True
            elif target < matrix[row][column]:
                column -= 1
            else:
                row += 1
        
        return False
    
    def searchMatrix2(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        return any(target in row for row in matrix)
    
    def searchMatrix3(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or len(matrix) == 0 or len(matrix[0]) == 0:
            return False
        
        def helper(nums, target):
            left, right = 0, len(nums) - 1
            
            while left <= right:
                # remember to use // 2
                middle = (left + right) // 2

                if target == nums[middle]:
                    return True
                elif target < nums[middle]:
                    right = middle - 1
                else:
                    left = middle + 1
            
            return False
        
        for row in matrix:
            if helper(row, target):
                return True
        
        return False