Value range search based on the Binary Search Method

1. Problem definition:

In an ordered array, search for all elements in the input array that meet the conditions of a value greater than a value and a value less than B. Where a and B may not be in the array.

2. Example description:

Input array eg: {, 33, 45, 49 }. If all the numbers greater than 10 and less than 30 are found, the output result is a subset of {11, 16, 31. If the query value is greater than 50 or less than 2, the output empty set is displayed.

3. linear scanning:

Complexity O (n) scanning from left to right to meet the condition range

4. Binary Search:

Complexity O (logn) binary search meets the condition range, two binary searches, determine the subarray range.

auto binary_search_left(const vector<int>& nums,int min,int max){     if(nums.empty())       return -1;     int left=0;     int right =nums.size()-1;     auto min_left=-1;     while(left<=right)     {         auto mid=(left+right)/2;         if(nums[mid]<min){             left=mid+1;             continue;         }         if(nums[mid]>max){             right=mid-1;             continue;         }            min_left=mid;            right=mid-1;     }     return min_left;}auto binary_search_right(const vector<int>& nums,int min,int max){     if(nums.empty())       return -1;     int left=0;     int right =nums.size()-1;     auto max_right=-1;     while(left<=right)     {         auto mid=(left+right)/2;         if(nums[mid]<min){             left=mid+1;//mid+1             continue;         }         if(nums[mid]>max){             right=mid-1;             continue;         }            max_right=mid;            left=mid+1;     }     return max_right;}

5. Where can I apply it?

For example, the tower anti-DDoS game has a lot of bullet detection, and the dual for loop traverses all bullets and monsters will have a lot of redundant detection graphics intersection calculation. So what should we do? Our space division technology sorts bullets or monsters by X axis coordinates. Only bullets or monsters within a certain range need collision detection (Graphic intersection calculation ). Fast sorting is adopted. The time complexity O (nlogn) is located in a sorted array to find the intervals that meet the conditions. The time complexity is O (logn) by means of binary search ). Although sorting and search operations are introduced, the cost model of sorting and search is only a comparison operation and a value assignment operation. What's better is that we greatly reduce collision detection (Graphic intersection calculation ).