You are given n numbers a1, a2, . . . , an. It takes constant time to check whether two numbers ai and aj are of the same value. The goal is to check whether more than half of the numbers have the same value. Design an O(n log n) time algorithm to solve the problem.写出详细的算法设计

算法设计：1.使用快速排序算法对数组a1，a2，...，an进行排序，时间复杂度O(nlog n)。2.从头到尾遍历排序后的数组，记录相同值的数字的个数。3.如果某个值的个数大于总个数的一半，那么此值就是该数组中超过一半的数字值。4.时间复杂度：O(n log n)。

相关问题

You are given n numbers a1, a2, . . . , an. It takes constant time to check whether two numbers ai and aj are of the same value. The goal is to check whether more than half of the numbers have the same value. Design an O(n log n) time algorithm to solve the problem.详细的分析运行时间

答：我们可以使用快速排序算法来解决这个问题。先对这n个数字进行排序，时间复杂度为O(n log n) 。排序完成后，可以在常数时间内判断相邻的数字是否相等。如果有大于一半的数字相等，那么排序后它们会聚集在一起。因此，只需要在排序结果中检查相邻的数字即可，从而完成对问题的求解。

An M × N matrix of real numbers is given. Find the amount of its rows whose values of elements are sorted in ascending order.

One way to solve this problem is to iterate through each row of the matrix and check if its elements are sorted in ascending order. This can be done by comparing each element to the previous one and checking if it is greater or equal to it. If all elements satisfy this condition, then the row is sorted in ascending order. We can count the number of rows that satisfy this condition and return the result.

Here is the Python code to implement this approach:

def count_sorted_rows(matrix):
count = 0
for row in matrix:
sorted_row = True
for i in range(1, len(row)):
if row[i] < row[i-1]:
sorted_row = False
break
if sorted_row:
count += 1
return count

# Example usage:
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
[3, 2, 1]
]
print(count_sorted_rows(matrix)) # Output: 2

In this example, the matrix has two sorted rows: [1, 2, 3] and [4, 5, 6]. The function returns 2 as the result.