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);*/