Title Description

Statistics the number of times a number appears in a sorted array.

Time limit: 1 second; Space limit: 32768K; Knowledge point: Array

Solving problems

Train of thought

Find the location index of the first identical number first, and then judge whether the number of consecutive numbers is identical from this position until different numbers appear and output the counting results.

# -*- coding:utf-8 -*-
class Solution:
    def GetNumberOfK(self, data, k):
        # write code here
        if data==[]: #Consider all possible scenarios
            return 0
        count = 0
        # If so, find the first index of the same number
        for i in range(len(data)):
            if data[i]==k:
                break
        # Loop until the next number is different
        while data[i]==k:
            count += 1
            i += 1
            # If it's the last one, jump out of the loop
            if i == len(data):
                break
        return count

Train of thought two

Use dichotomy to find the first and last position of k.

# -*- coding:utf-8 -*-
class Solution:
    def GetNumberOfK(self, data, k):
        # write code here
        # Binary search locates the first and last position of k
        lower = self.getLower(data,k)
        upper = self.getUpper(data,k)
        return upper-lower+1
    # Get the subscript for the first occurrence of k
    def getLower(self, data, k):
        start = 0
        end = len(data)-1
        mid = (start + end) // 2
        while start <= end:
            if data[mid] < k: #Pay attention to differences
                start = mid + 1
            else:
                end = mid - 1
            mid = (start + end) // 2
        return start
    # Get the subscript for the last occurrence of k
    def getUpper(self, data, k):
        start = 0
        end = len(data)-1
        mid = (start + end) // 2
        while start <= end:
            if data[mid] <= k: #Pay attention to differences
                start = mid + 1
            else:
                end = mid - 1
            mid = (start + end) // 2
        return end