436. Find Right Interval ------M

 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; } };