[LeetCode] 436. Find Right Interval

[grandyang]

 

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. You may assume none of these intervals have the same start point.

 

Example 1:

Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

 

Example 2:

Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.

 

Example 3:

Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.

 

这道题给了我们一堆区间，让我们找每个区间的最近右区间，要保证右区间的 start 要大于等于当前区间的 end，由于区间的顺序不能变，所以我们不能给区间排序，我们需要建立区间的 start 和该区间位置之间的映射，由于题目中限定了每个区间的 start 都不同，所以不用担心一对多的情况出现。然后我们把所有的区间的 start 都放到一个数组中，并对这个数组进行降序排序，那么 start 值大的就在数组前面。然后我们遍历区间集合，对于每个区间，我们在数组中找第一个小于当前区间的 end 值的位置，如果数组中第一个数就小于当前区间的 end，那么说明该区间不存在右区间，结果 res 中加入-1；如果找到了第一个小于当前区间 end 的位置，那么往前推一个就是第一个大于等于当前区间 end 的 start，我们在 HashMap 中找到该区间的坐标加入结果 res 中即可，参见代码如下：

 

解法一：

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        vector<int> res, starts;
        unordered_map<int, int> m;
        for (int i = 0; i < intervals.size(); ++i) {
            m[intervals[i][0]] = i;
            starts.push_back(intervals[i][0]);
        }
        sort(starts.rbegin(), starts.rend());
        for (auto interval : intervals) {
            int i = 0;
            for (; i < starts.size(); ++i) {
                if (starts[i] < interval[1]) break;
            }
            res.push_back((i > 0) ? m[starts[i - 1]] : -1);
        }
        return res;
    }
};

 

上面的解法可以进一步化简，我们可以利用 STL 的 lower_bound 函数来找第一个不小于目标值的位置，这样也可以达到我们的目标，参见代码如下：

 

解法二：

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        vector<int> res;
        map<int, int> m;
        for (int i = 0; i < intervals.size(); ++i) {
            m[intervals[i][0]] = i;
        }
        for (auto interval : intervals) {
            auto it = m.lower_bound(interval[1]);
            if (it == m.end()) res.push_back(-1);
            else res.push_back(it->second);
        }
        return res;
    }
};

 

Github 同步地址：

#436

 

类似题目：

Non-overlapping Intervals

Data Stream as Disjoint Intervals 

Insert Interval

Merge Intervals

 

参考资料：

https://leetcode.com/problems/find-right-interval/

https://leetcode.com/problems/find-right-interval/discuss/91819/C%2B%2B-map-solution

https://leetcode.com/problems/find-right-interval/discuss/91789/Java-clear-O(n-logn)-solution-based-on-TreeMap

 

LeetCode All in One 题目讲解汇总(持续更新中...)

[Sahilamin219]

Another approach without any STL is :
store hashed intervals with there indexes.
You can store both starting and ending intervals and short them .
After that just do a liner scan for each ending intervals .
Time:N*log(N)
Space:O(N)
Code