英文:
find the index of very first element that is greater than first limit but less than second limit efficiently
问题
I need to find the index of the very first element in a sorted std::vector<int> that is greater than the first limit but less than the second limit efficiently.
#include <iostream>
#include <vector>
int main() {
const int a = 2, b = 4; // a is always less than b
const std::vector<int> vec {1, 2, 3, 4, 5};
// an efficient algo that will return an iterator to the
// first element in vec that is greater than a but less than b
int ans = -1;
for (const auto x : vec) {
if (x > a && x < b) {
ans = x;
break;
}
}
std::cout << ans << std::endl;
}
英文:
I need to find the index of very first element in a sorted std::vector<int> that is greater than first limit but less than second limit efficiently.
#include <iostream>
#include <vector>
int main() {
const int a = 2, b = 4; // a is always less than b
const std::vector<int> vec {1, 2, 3, 4, 5};
// an efficient algo that will return an iterator to the
// first element in vec that is greater than a but less than b
int ans = -1;
for (const auto x : vec) {
if (x > a && x < b) {
ans = x;
break;
}
}
std::cout << ans << std::endl;
}
答案1
得分: 2
使用 std::upper_bound 函数：
返回一个迭代器，指向范围
[first, last)中第一个使得value < element（或comp(value, element)）为true（即严格大于）的元素，如果没有找到这样的元素，则返回last。
auto it = std::upper_bound(vec.begin(), vec.end(), a);
// 如果找到的值不小于 `b`，则将 `it` 设置为结束迭代器
if (it != vec.end() && not(*it < b)) it = vec.end();
英文:
Use std::upper_bound:
> Returns an iterator pointing to the first element in the range [first, last) such that value < element (or comp(value, element)) is true (i.e. strictly greater), or last if no such element is found.
auto it = std::upper_bound(vec.begin(), vec.end(), a);
// set `it` to the end iterator if the found value is not less than `b`
if(it != vec.end() && not(*it < b)) it = vec.end();
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