[LeetCode] 436. Find Right Interval 找右區間

 

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.

 

這道題給了我們一堆區間，讓我們找每個區間的最近右區間，要保證右區間的 start 要大於等於當前區間的 end，由於區間的順序不能變，所以我們不能給區間排序，我們需要建立區間的 start 和該區間位置之間的映射，由於題目中限定了每個區間的 start 都不同，所以不用擔心一對多的情況出現。然后我們把所有的區間的 start 都放到一個數組中，並對這個數組進行降序排序，那么 start 值大的就在數組前面。然后我們遍歷區間集合，對於每個區間，我們在數組中找第一個小於當前區間的 end 值的位置，如果數組中第一個數就小於當前區間的 end，那么說明該區間不存在右區間，結果 res 中加入-1；如果找到了第一個小於當前區間 end 的位置，那么往前推一個就是第一個大於等於當前區間 end 的 start，我們在 HashMap 中找到該區間的坐標加入結果 res 中即可，參見代碼如下：

 

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;
    }
};

 

上面的解法可以進一步化簡，我們可以利用 STL 的 lower_bound 函數來找第一個不小於目標值的位置，這樣也可以達到我們的目標，參見代碼如下：

 

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：

https://github.com/grandyang/leetcode/issues/436

 

類似題目：

Non-overlapping Intervals

Data Stream as Disjoint Intervals 

Insert Interval

Merge Intervals

 

參考資料：

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

 

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