You are given an array of intervals
, where intervals[i] = [starti, endi]
and each starti
is unique.
The right interval for an interval i
is an interval j
such that startj
>= endi
and startj
is minimized.
Return an array of right interval indices for each interval i
. If no right interval exists for interval i
, then put -1
at index i
.
Example 1:
Input: intervals = [[1,2]] Output: [-1] Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: intervals = [[3,4],[2,3],[1,2]] Output: [-1,0,1] Explanation: There is no right interval for [3,4]. The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3. The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:
Input: intervals = [[1,4],[2,3],[3,4]] Output: [-1,2,-1] Explanation: There is no right interval for [1,4] and [3,4]. The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
Constraints:
1 <= intervals.length <= 2 * 104
intervals[i].length == 2
-106 <= starti <= endi <= 106
- The start point of each interval is unique.
A:
就是先根据每个interval 的start 排序,然后再逐次往下一个找。
这个题的tricky地方是:要做好各种mapping . 我犯错的一点儿就是,一开始没有把query interval 的index map好, 而直接用了push_back. 哎,恰好25分钟做完,还是不太够
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | class Solution { public: vector<int> findRightInterval(vector<vector<int>>& intervals) { unordered_map<int,int> M;// start_i --> its idex int n = intervals.size(); for(int i =0; i<n;i++){ M[intervals[i][0]] = i; } // Using assignment operator to copy one vector to other sort(intervals.begin(), intervals.end()); vector<int> res(n,0); for(int i =0;i< n;i++){ int end = i+1; while(end<n && intervals[end][0] < intervals[i][1]){ end++; } res[M[intervals[i][0]]] = (end==n)?-1 :M[intervals[end][0]]; } return res; } }; |
No comments:
Post a Comment