Brush Find prove safety offer two-dimensional array of records in question

1 Title Description

(The length of each one-dimensional array of same), each row from left to right in order of ascending sort, to sort each column in a two-dimensional array in order of increasing from top to bottom. A complete function, enter such a two-dimensional array and an integer, it is determined whether the array contains the integer.

2 violent problem-solving

2.1 Analysis

One by one through the array, it returns true if found, searched the entire array can not be found, it returns false.

2.2 Code

class Solution:
    # array 二维列表
    def Find(self, target, array):
        for i in array:
            for j in i:
                if j == target:
                    return True
        return False

2.3 Complexity

time complexity: $O (n ^ 2)$
space complexity: $O (1)$

3 from the top right corner to start looking

3.1 Analysis

The conditions given, the array is incremented from left to right, top to bottom incremented;
that is the right-most element in each row for the row maximum value;
Analyzing:
This value is equal to target, return true;
this value is less than the target, excluded the row, down one line;
this value is greater than the target, the column is excluded, a left;

3.2 Code

class Solution:
    # array 二维列表
    def Find(self, target, array):
        rows = len(array)
        cols = len(array[0])
        
        i = 0
        j = cols - 1
        while i<rows and j>=0:
            if array[i][j] == target:
                return True
            elif array[i][j] < target:
                i = i + 1
            elif array[i][j] > target:
                j = j - 1
        return False

3.3 Complexity

Time complexity: O (rows + cols)
Complexity Space: O (1)