leetcode 2080. 区间内查询数字的频率

题目如下

数据范围

示例

这题十分有意思一开始我想对每个子数组排序二分结果超时了。
转换思路：我们可以提前把每个数字出现的位置先记录下来形成集合，
然后拿着left和right利用二分查找看看left和right是不是在集合里然后做一个相减就出答案了。

通过代码

class RangeFreqQuery {
public:
   unordered_map<int,vector<int>> map;
   //unordered_map<string,int> mapc;
    RangeFreqQuery(vector<int>& arr) {
        int n = arr.size();
        for(int i = 0;i < n;i++){
           if(map.count(arr[i]) == 1)map[arr[i]].emplace_back(i);
           else{
            map[arr[i]] = vector<int>{i};
           }
        }
    }
      int findlow(vector<int>& nums,int target) {
        int n = nums.size();
        int l = 0, r = n - 1;
        int mid;
        while (l < r) {
            mid = (l + r) / 2;
            if (nums[mid] >= target) {
                r = mid;
            } else{
                l = mid + 1;
            }
        }
        if(nums[l] < target)return -1;
     
        return l;
    }
    int findhigh(vector<int>& nums,int target) {

        int n = nums.size();
        int l = 0, r = n - 1;
        int mid;
        while (l < r) {
            mid = (l + r + 1) / 2;
            if (nums[mid] > target) {
                r = mid - 1;
            } else{
                l = mid;
            }
        }
        if(nums[l] > target)return -1;
        return l;
    }
    int query(int left, int right, int value) {
      //  string a = to_string(left) + "+" + to_string(right) + "+" + to_string(value);
      //  if(mapc.count(a) == 1)return mapc[a];
        if(map.count(value) == 0)return 0;
        vector<int>& v = map[value];
        int l = findlow(v,left);
        int r = findhigh(v,right);
        if(l != -1 && r != -1){
      //      mapc[a] = r - l + 1;
            return r - l + 1;
            }
        return 0;
    }
};

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery* obj = new RangeFreqQuery(arr);
 * int param_1 = obj->query(left,right,value);
 */