74. Search a 2D Matrix 240. Search a 2D Matrix II python

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最新推荐文章于 2020-12-22 13:11:29 发布

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Write an efficient algorithm that searches for a value in an mx 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.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

题意：给一个二维list，从左到右，从上到下拍好序的，然后判断target是否在二维list中

思路：1：把全部的值放到set中，然后用in判断，缺点：占用内存较大

2：判断是否大于当前列第一个值，小于最后一个值，是：用in判断是否存在，不是：进入下一列。优点：节省空间

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        '''
        if(matrix==[]):
            return False
        ma_list=set()
        row=len(matrix)
        column=len(matrix[0])
        for i in range(row):
            for j in range(column):
                ma_list.add(matrix[i][j])
        if(target in ma_list):
            return True
        else:
            return False
        '''
        if(matrix==[] or matrix==[[]]):
            return False
        row=len(matrix)
        column=len(matrix[0])
        for i in range(row):
            if(target<matrix[i][0]):
                return False
            elif(target<=matrix[i][column-1]):
                if(target in matrix[i]):
                    return True
                else:
                    return False
            else:
                continue
        return False

Write an efficient algorithm that searches for a value in an mx 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.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

思路：从左下角判断，如果比他大就向上，小就向右

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        '''
        if(matrix==[]):
            return False
        ma_list=set()
        row=len(matrix)
        column=len(matrix[0])
        for i in range(row):
            for j in range(column):
                ma_list.add(matrix[i][j])
        if(target in ma_list):
            return True
        else:
            return False
        '''
        if(matrix==[] or matrix==[[]]):
            return False
        row=len(matrix)
        column=len(matrix[0])
        for i in range(row):
            if(target<matrix[i][0]):
                return False
            elif(target<=matrix[i][column-1]):
                if(target in matrix[i]):
                    return True
                else:
                    return False
            else:
                continue
        return False