53-01 数字在排序数组中出现的次数

题目

统计一个数字在排序数组中出现的次数。例如输入排序数组{1,2,3,3,3,3,4,5}和数字3，由于3在这个数组中出现了4次，因此输出4。

牛客网 OJ
AcWing OJ

C 语言题解

二分查找，找到第一次出现的位置以及最后出现的位置：

int GetFirstK(const int* data, int length, int k, int start, int end);
int GetLastK(const int* data, int length, int k, int start, int end);

int GetNumberOfK(const int* data, int length, int k)
{
	int number = 0;

	if (data != nullptr && length > 0)
	{
		int first = GetFirstK(data, length, k, 0, length - 1);
		int last = GetLastK(data, length, k, 0, length - 1);

		// 最后一个k的位置与第一个k的位置作差+1即为数字k出现的次数
		if (first > -1 && last > -1)
			number = last - first + 1;
	}

	return number;
}

// 找到数组中第一个k的下标。如果数组中不存在k，返回-1
int GetFirstK(const int* data, int length, int k, int start, int end)
{
	if (start > end)
		return -1;

	int middleIndex = (start + end) / 2;
	int middleData = data[middleIndex];

	if (middleData == k)
	{
		if ((middleIndex > 0 && data[middleIndex - 1] != k)
			|| middleIndex == 0)
			return middleIndex;
		else
			end = middleIndex - 1;
	}
	else if (middleData > k)
		end = middleIndex - 1;
	else
		start = middleIndex + 1;

	return GetFirstK(data, length, k, start, end);
}

// 找到数组中最后一个k的下标。如果数组中不存在k，返回-1
int GetLastK(const int* data, int length, int k, int start, int end)
{
	if (start > end)
		return -1;

	int middleIndex = (start + end) / 2;
	int middleData = data[middleIndex];

	if (middleData == k)
	{
		if ((middleIndex < length - 1 && data[middleIndex + 1] != k)
			|| middleIndex == length - 1)
			return middleIndex;
		else
			start = middleIndex + 1;
	}
	else if (middleData < k)
		start = middleIndex + 1;
	else
		end = middleIndex - 1;

	return GetLastK(data, length, k, start, end);
}

C++ 题解

class Solution {
public:
    int GetNumberOfK(vector<int> data ,int k) {
        int first = getFirstK(data,k);
        int last = getLastK(data,k);
        
        return last - first + 1;        
    }
    
    int getFirstK(vector<int>data,int k)
    {
        int length = data.size();
        int left = 0;
        int right = length - 1;
        int mid = (left+right)/2;
        
        while(left <= right)
        {
            // 这里需要带等于号
            if (data[mid] >= k)
            {
                right = mid - 1;
            }
            else
            {
                left = mid + 1;
            }
            mid = (right+left) / 2;            
        }
        
        return left;
    }
                  
    int getLastK(vector<int>data,int k)
    {
        int length = data.size();
        int left = 0;
        int right = length - 1;
        int mid = (left+right) /2;
        
        // 这里需要带等于号
        while(left <= right)
        {
            if (data[mid] <= k)
            {
                left = mid + 1;
            }
            else
            {
                right = mid - 1;
            }
            mid = (right+left) / 2;            
        }
        
        return right;        
    }
};

python 题解

# -*- coding:utf-8 -*-
class Solution:
    def GetNumberOfK(self, data, k):
        # write code here
        first = self.getFirstK(data,k)
        last = self.getLastK(data,k)
        
        return last - first + 1
    
    def getFirstK(self,num,k):
        left,right = 0,len(num)-1
        mid = (left + right) // 2
        
        while left <= right:
            if num[mid] >= k:
                right = mid -1
            else:
                left = mid +1
            mid = (left + right) // 2
        
        return left
    
    def getLastK(self,num,k):
        left,right = 0,len(num)-1
        mid = (left + right) // 2
        
        while left <= right:
            if num[mid] <= k:
                left = mid + 1
            else:
                right = mid - 1 
            mid = (left + right) // 2
        
        return right

注意：

• python除法采用//