djasper added a comment. There might be an even easier algorithm:
What if we put all start and end points into a single sorted list. Identical start points are sorted by decreasing end points and identical end points are sorted by increasing start points. Now do a single sweep and simply count the total number of starts vs. the total number of ends, call this C (if you encounter a start point, ++C, if you encounter an end point, --C). If you encounter a start-point and C != 0, mark the interval and corresponding Fixit as inapplicable. Similarly if you encounter an end point and C != 1, mark the fixit as inapplicable. Basically, this does a single sweep over all fixits and marks the ones as bad that have either their start point or their endpoint within a different interval. (Take this as an additional idea, if you have an existing algorithm that you like better, I am fine with keeping that. Just wanted to give my additional thoughts). http://reviews.llvm.org/D13516 _______________________________________________ cfe-commits mailing list cfe-commits@lists.llvm.org http://lists.llvm.org/cgi-bin/mailman/listinfo/cfe-commits
