C 解法, 执行用时: 2ms, 内存消耗: 324KB, 提交时间: 2021-06-06

/**
 * 
 * @param data int整型一维数组 
 * @param dataLen int data数组长度
 * @param k int整型 
 * @return int整型
 */



int bin_serch(const int* data, int dataLen, int k) 
{
    int left=0;
    int right=dataLen-1;
    
    while(left<right)
    {
        int avarage=(left+right)/2;
        if(k==data[avarage])return avarage;
        else if(k>data[avarage]) 
        {
            left=avarage+1;
        }
        else
        {
            right=avarage-1;
        }
    }
    return -1;
}










int GetNumberOfK(int* data, int dataLen, int k ) {
 //int different=0;
 
    int count=0;
    int left=bin_serch(data,dataLen,  k) ;
    int right=left+1;
    while(data[left]==k&&left>=0)
    {
        count++;
        left--;
    }
        while(data[right]==k&&right<dataLen)
    {
        count++;
        right++;
    }
    
    return count;
    // write code here
}

C++ 解法, 执行用时: 2ms, 内存消耗: 352KB, 提交时间: 2021-04-25

class Solution {
public:
    int GetNumberOfK(vector<int> data ,int k) {
        int lbound = 0, rbound = 0;
        // 二分找到上届和下届
        
        // 找下届
        int left = 0;
        int right = data.size();
        int p1;
        while(left < right)
        {
            p1 = (left + right) / 2;
            
            if (data[p1] >= k)    // 往左侧
                right = p1;
            else   // 右侧
                left = p1+1;
        }
        lbound = left;
        // 找下届
        left = 0;
        right = data.size();
        int p2;
        while(left < right)
        {
            p2 = (left + right) / 2;
            
            if (data[p2] > k)    // 往左侧
                right = p2;
            else   // 右侧
                left = p2+1;
        }
        rbound = left;
        
//        if (data[p1] != k)
//            return 0;
        return rbound - lbound;
    }
};

C 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-06-09

/**
 * 
 * @param data int整型一维数组 
 * @param dataLen int data数组长度
 * @param k int整型 
 * @return int整型
 */
int GetNumberOfK(int* data, int dataLen, int k ) {
    int count =0;
    for(int i = 0;i<dataLen;i++)
    {
        if (*(data+i) == k)
            count+=1;
        
    }
    return count;
    // write code here
}

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-05-25

class Solution {
public:
    int GetNumberOfK(vector<int> data ,int k) 
    {
        if(k<data[0]) return 0;
        int num=0;
        for(int i=0;i<data.size();i++)
        {
            if(k==data[i])
            {
                num++;
            }
            else if(k<data[i]) break;
        }
        return num;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2021-05-19

class Solution {
public:
    int GetNumberOfK(vector<int> data ,int k) {
        int l=0,r = data.size()-1;
        int flag=0;
        int mid;
        while(l<=r){
            mid = (r-l)/2+l;
            if(data[mid]>k){
                r = mid-1;
            }else if(data[mid]<k){
                l = mid+1;
            }else if(data[mid]==k){
                flag=1;
                break;
            }
        }
        if(flag==0) return 0;
        int j = mid-1,i = mid +1;
        while((j>=0&&data[j]==k)||(i<=data.size()-1&&data[i]==k)){
            if(j>=0&&data[j]==k){
                flag++;
                j--;
            }
            if(i<=data.size()-1&&data[i]==k){
                flag++;
                i++;
            }
        }
        return flag;
    }
};