240. Search a 2D Matrix II

solution

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.
For 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.

可能在多个分块中

solution

分块索引直接引申出的思维方式

时间复杂度 O(m*n)

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        for i in matrix:
            if target > i[-1]:
                continue
            if target in i:
                return True
        return False

O(m+n)

利用了题目一个隐藏的特性，每一个点都比它上面左边的点大

class Solution:
    # @param {integer[][]} matrix
    # @param {integer} target
    # @return {boolean}
    def searchMatrix(self, matrix, target):
        if matrix:
            row,col,width=len(matrix)-1,0,len(matrix[0])
            while row>=0 and col<width:
                if matrix[row][col]==target:
                    return True
                elif matrix[row][col]>target:
                    row=row-1
                else:
                    col=col+1
            return False