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，我们在哈希表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
     
       findRightInterval(vector
      
       & intervals) {        vector
       
         res, v;        unordered_map
        
          m;        for (int i = 0; i < intervals.size(); ++i) {            m[intervals[i].start] = i;            v.push_back(intervals[i].start);        }        sort(v.begin(), v.end(), greater
         
          ());        for (auto a : intervals) {            int i = 0;            for (; i < v.size(); ++i) {                if (v[i] < a.end) break;            }            res.push_back((i > 0) ? m[v[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
     
       findRightInterval(vector
      
       & intervals) {        vector
       
         res;        map
        
          m;        for (int i = 0; i < intervals.size(); ++i) {            m[intervals[i].start] = i;        }        for (auto a : intervals) {            auto it = m.lower_bound(a.end);            if (it == m.end()) res.push_back(-1);            else res.push_back(it->second);        }        return res;    }};
        
       
      
     

 

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