反正不想用递归
但是一说是排序的 最好又用二分查找,这样比较快;
方法一 :找出重复数字出现第一次的坐标。以及出现最后一次的坐标;
链接:https://www.nowcoder.com/questionTerminal/70610bf967994b22bb1c26f9ae901fa2?f=discussion
来源:牛客网
class Solution {
public:
int GetNumberOfK(vector<int> data ,int k) {
int lower = getLower(data,k);
int upper = getUpper(data,k);
return upper - lower + 1;
}
//获取k第一次出现的下标
int getLower(vector<int> data,int k){
int start = 0,end = data.size()-1;
int mid = (start + end)/2;
while(start <= end){
if(data[mid] < k){
start = mid + 1;
}else{
end = mid - 1;
}
mid = (start + end)/2;
}
return start;
}
//获取k最后一次出现的下标
int getUpper(vector<int> data,int k){
int start = 0,end = data.size()-1;
int mid = (start + end)/2;
while(start <= end){
if(data[mid] <= k){
start = mid + 1;
}else{
end = mid - 1;
}
mid = (start + end)/2;
}
return end;
}
};
方法二: 利用一点小技巧
链接:https://www.nowcoder.com/questionTerminal/70610bf967994b22bb1c26f9ae901fa2?f=discussion
来源:牛客网
//因为data中都是整数,所以可以稍微变一下,不是搜索k的两个位置,而是搜索k-0.5和k+0.5
//这两个数应该插入的位置,然后相减即可。
class Solution {
public:
int GetNumberOfK(vector<int> data ,int k) {
return biSearch(data, k+0.5) - biSearch(data, k-0.5) ;
}
private:
int biSearch(const vector<int> & data, double num){
int s = 0, e = data.size()-1;
while(s <= e){
int mid = (e - s)/2 + s;
if(data[mid] < num)
s = mid + 1;
else if(data[mid] > num)
e = mid - 1;
}
return s;
}
};