Highest Frequency number in given intervals.

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本文介绍了一种通过暴力搜索的方法来找到一个整数，该整数能够落在给定区间集合中的尽可能多的区间内。示例代码展示了如何确定这个整数，并提供了具体的实现细节。

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find a number in the given intervals, make this point fall in as many intervals as possible.
for example: [2, 3], [3, 5], [4, 5], [1, 5], [1, 2]
return 4 or 5

Tips: using brute force.

#include <vector>
#include <iostream>
#include <climits>
using namespace std;

// question 2, using brute force.
struct Interval {
  int start;
  int end;
  Interval(int a, int b) : start(a), end(b) {}
};

int pointInMaxInterval(vector<Interval>& nums) {
  int minValue = INT_MAX;
  int maxValue = INT_MIN;
  for(int i = 0; i < nums.size(); ++i) {
    minValue = min(minValue, nums[i].start);
    maxValue = max(maxValue, nums[i].end);
  }
  struct nodes {
    int val;
    int count;
    nodes(int a, int b) : val(a), count(b) {}
  };
  nodes maxCount(0, 0);
  for(int i = minValue; i <= maxValue; ++i) {
    int count = 0;
    for(int j = 0; j < nums.size(); ++j) {
      if(nums[j].start <= i && nums[j].end >= i) {
        count++;
      }
    }
    if(count > maxCount.count) {
      maxCount.val = i;
      maxCount.count = count;
    }
  }
  return maxCount.val;
}

int main(void) {
  vector<Interval> nums{{8, 10}, {3, 5}, {4, 5}, {1, 5}, {1, 2}};
  cout << pointInMaxInterval(nums) << endl;
}