[leetcode] 240. Search a 2D Matrix II

题目大意

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

240. Search a 2D Matrix II

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.

编写一个高效的算法，从一个m × n矩阵中寻找一个值。矩阵具有如下性质：

每一行的整数从左向右递增
每一列的整数从上往下递增

测试样例见题目描述。

解题思路

解法一：从矩阵的右上角(屏幕坐标系）开始，执行两重循环；外循环递增枚举每行，内循环递减枚举列

class Solution(object):
    def searchMatrix(self, matrix, target):  # 48ms
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not len(matrix) or not len(matrix[0]):
            return False
        m, n = len(matrix), len(matrix[0])
        r, c = 0, n - 1
        while r < m and c >= 0:
            if matrix[r][c] == target:
                return True
            elif matrix[r][c] > target:
                c -= 1
            else:
                r += 1
        return False

算法复杂度：O(m + n)

类似思路的实现：

class Solution(object):        
    def searchMatrix(self, matrix, target):  # 48ms
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not len(matrix) or not len(matrix[0]):
            return False
        y = len(matrix[0]) - 1
        for x in range(len(matrix)):
            while y and matrix[x][y] > target:
                y -= 1
            if matrix[x][y] == target:
                return True
        return False

解法二：循环枚举行，二分查找列

class Solution(object):       
    def searchMatrix(self, matrix, target):  # 二分查找 148ms
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not len(matrix) or not len(matrix[0]):
            return False
        y = len(matrix[0]) - 1
        
        def binary_search(nums, low, high):
            while low <= high:
                mid = (low + high) / 2
                if nums[mid] > target:
                    high = mid - 1
                else:
                    low = mid + 1
            return high

算法复杂度：O(m * logn)

参考：

http://bookshadow.com/weblog/2015/07/23/leetcode-search-2d-matrix-ii/

https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/183609/An-intelligible-Python-solution-beats-99.64-48ms