leetcode 【Search a 2D Matrix 】python 实现

题目：

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

代码：oj测试通过 Runtime: 75 ms

class Solution:
    # @param matrix, a list of lists of integers
    # @param target, an integer
    # @return a boolean
    def searchInline(self, line, target):
        start = 0
        end = len(line)-1
        while start <= end :
            if start == end :
                return [False,True][line[start]==target]
            if start+1 == end :
                if line[start]==target or line[end]==target :
                    return True
                else:
                    return False
            mid=(start+end)/2
            if line[mid]==target :
                return True
            elif line[mid]>target :
                end = mid-1
            else :
                start = mid+1

    def searchMatrix(self, matrix, target):
        if len(matrix) == 0 :
            return False
        
        if len(matrix) == 1 :
            return self.searchInline(matrix[0], target)
            
        if len(matrix) == 2 :
            return self.searchInline([matrix[1],matrix[0]][matrix[1][0]>target], target)
            
        start = 0
        end = len(matrix)-1
        while start <= end :
            if start == end:
                return self.searchInline(matrix[start],target)
            if start+1 == end:
                if matrix[start][0] <= target and matrix[end][0] > target:
                    return self.searchInline(matrix[start],target)
                if matrix[end][0] < target :
                    return self.searchInline(matrix[end],target)
            mid = (start+end+1)/2
            if matrix[mid][0] <= target and matrix[mid+1][0] > target:
                return self.searchInline(matrix[mid],target)
            elif matrix[mid][0] > target :
                end = mid-1
            else :
                start = mid+1

思路：

先按行二分查找，再按列二分查找。

代码写的比较繁琐。