Contained Intervals - Csacademy


https://csacademy.com/contest/interview-archive/task/contained-intervals/
You should implement a function that takes as argument an array of N intervals. For each interval i you are given its end points (l, r). We say that an interval i is contained by another interval j if l_j \leq l_i and r_i \leq r_j. Return the number of intervals that are contained by at least one other interval.

Desired solution

This problem admits a variety of O(N\log N) solutions. Can you find the simplest one?

Standard input

The template code reads on the first line a single integer N.
Each of the next N lines contains two integers l and r representing the end points of an interval.

Standard output

The template calls your function and outputs the result.
InputOutput
5
1 6
4 6
4 10
5 6
1 3
3
2
1 2
1 2
2
https://zhuanlan.zhihu.com/p/26657786
给一组interval,判断有多少个interval是被至少一个别的interval所包含的(包含的定义:a包含b,就是 a.start <= b.start, a.end >= b.end)。
这道题的思路就是,先按照Interval的start升序排序,对于start相同的interval,按照interval的end的降序排序。这样子可以保证在排序后,对于每一个interval,如果它被另一个interval所包含,那么那个包含它的interval一定在它之前。这是排序的结果所造成的。同时因为排序,每一个interval的start一定大于等于之前的所有interval,所以只要之前有任意一个interval的end大于等于这个interval的end,那么就满足了包含的条件。于是我们就只需要在遍历排序过后的数组时,记录下出现过的最大的end,然后和当下的interval的end作比较。唯一需要注意的一点是,如果前后两个interval的start和end完全相等,那么意味着这两个interval同时被对方所包含,在这种情况下,我们不能漏算前一个interval的情况。因此我们每次都还要比较一下每个interval和后一个interval是否相等。来看一下我的代码,值得解释的是,因为对于排序过后的第一个元素,currentmaxR的比较条件必然成立,所以在初始化res的时候要考虑是否初始化成-1来抵消这次误加。
struct Interval {
    int l, r;
};
bool operator==(const Interval& lhs, const Interval& rhs) {
    return (lhs.l == rhs.l) && (lhs.r == rhs.r);
}
bool myfunction (Interval i,Interval j) {
    if(i.l != j.l)  return (i.l < j.l);
    return (i.r > j.r); 
}
int containedIntervals(vector<Interval>& intervals) {
    if(intervals.empty())   return 0;
    sort(intervals.begin(), intervals.end(), myfunction);
    int currentMaxR = 0;
    int res = intervals.size() > 1 && intervals[0] == intervals[1] ? 0 : -1;
    currentMaxR = intervals[0].r;
    int i = 0;
    for(;i < intervals.size()-1; ++i) {
        if(intervals[i] == intervals[i+1] || intervals[i].r <= currentMaxR)  ++res;
        currentMaxR = max(currentMaxR, intervals[i].r);
    }
    if(intervals[i].r <= currentMaxR)   ++res;
    return res;
}

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