| Guido van Rossum | 4acc25b | 2000-02-02 15:10:15 +0000 | [diff] [blame] | 1 | """Bisection algorithms.""" |
| Guido van Rossum | 4e16098 | 1992-09-02 20:43:20 +0000 | [diff] [blame] | 2 | |
| Tim Peters | 36cdad1 | 2000-12-29 02:06:45 +0000 | [diff] [blame] | 3 | def insort_right(a, x, lo=0, hi=None): |
| 4 | """Insert item x in list a, and keep it sorted assuming a is sorted. |
| 5 | |
| 6 | If x is already in a, insert it to the right of the rightmost x. |
| 7 | |
| 8 | Optional args lo (default 0) and hi (default len(a)) bound the |
| 9 | slice of a to be searched. |
| 10 | """ |
| 11 | |
| Guido van Rossum | 4acc25b | 2000-02-02 15:10:15 +0000 | [diff] [blame] | 12 | if hi is None: |
| 13 | hi = len(a) |
| 14 | while lo < hi: |
| 15 | mid = (lo+hi)/2 |
| 16 | if x < a[mid]: hi = mid |
| 17 | else: lo = mid+1 |
| 18 | a.insert(lo, x) |
| Guido van Rossum | 4e16098 | 1992-09-02 20:43:20 +0000 | [diff] [blame] | 19 | |
| Tim Peters | 36cdad1 | 2000-12-29 02:06:45 +0000 | [diff] [blame] | 20 | insort = insort_right # backward compatibility |
| Guido van Rossum | 4e16098 | 1992-09-02 20:43:20 +0000 | [diff] [blame] | 21 | |
| Tim Peters | 36cdad1 | 2000-12-29 02:06:45 +0000 | [diff] [blame] | 22 | def bisect_right(a, x, lo=0, hi=None): |
| 23 | """Return the index where to insert item x in list a, assuming a is sorted. |
| 24 | |
| 25 | The return value i is such that all e in a[:i] have e <= x, and all e in |
| 26 | a[i:] have e > x. So if x already appears in the list, i points just |
| 27 | beyond the rightmost x already there. |
| 28 | |
| 29 | Optional args lo (default 0) and hi (default len(a)) bound the |
| 30 | slice of a to be searched. |
| 31 | """ |
| 32 | |
| Guido van Rossum | 4acc25b | 2000-02-02 15:10:15 +0000 | [diff] [blame] | 33 | if hi is None: |
| 34 | hi = len(a) |
| 35 | while lo < hi: |
| 36 | mid = (lo+hi)/2 |
| 37 | if x < a[mid]: hi = mid |
| 38 | else: lo = mid+1 |
| 39 | return lo |
| Tim Peters | 36cdad1 | 2000-12-29 02:06:45 +0000 | [diff] [blame] | 40 | |
| 41 | bisect = bisect_right # backward compatibility |
| 42 | |
| 43 | def insort_left(a, x, lo=0, hi=None): |
| 44 | """Insert item x in list a, and keep it sorted assuming a is sorted. |
| 45 | |
| 46 | If x is already in a, insert it to the left of the leftmost x. |
| 47 | |
| 48 | Optional args lo (default 0) and hi (default len(a)) bound the |
| 49 | slice of a to be searched. |
| 50 | """ |
| 51 | |
| 52 | if hi is None: |
| 53 | hi = len(a) |
| 54 | while lo < hi: |
| 55 | mid = (lo+hi)/2 |
| 56 | if a[mid] < x: lo = mid+1 |
| 57 | else: hi = mid |
| 58 | a.insert(lo, x) |
| 59 | |
| 60 | |
| 61 | def bisect_left(a, x, lo=0, hi=None): |
| 62 | """Return the index where to insert item x in list a, assuming a is sorted. |
| 63 | |
| 64 | The return value i is such that all e in a[:i] have e < x, and all e in |
| 65 | a[i:] have e >= x. So if x already appears in the list, i points just |
| 66 | before the leftmost x already there. |
| 67 | |
| 68 | Optional args lo (default 0) and hi (default len(a)) bound the |
| 69 | slice of a to be searched. |
| 70 | """ |
| 71 | |
| 72 | if hi is None: |
| 73 | hi = len(a) |
| 74 | while lo < hi: |
| 75 | mid = (lo+hi)/2 |
| 76 | if a[mid] < x: lo = mid+1 |
| 77 | else: hi = mid |
| 78 | return lo |