|
| 1 | +/* |
| 2 | + Author: King, wangjingui@outlook.com |
| 3 | + Date: Dec 20, 2014 |
| 4 | + Problem: Search for a Range |
| 5 | + Difficulty: Medium |
| 6 | + Source: https://oj.leetcode.com/problems/search-for-a-range/ |
| 7 | + Notes: |
| 8 | + Given a sorted array of integers, find the starting and ending position of a given target value. |
| 9 | +
|
| 10 | + Your algorithm's runtime complexity must be in the order of O(log n). |
| 11 | +
|
| 12 | + If the target is not found in the array, return [-1, -1]. |
| 13 | +
|
| 14 | + For example, |
| 15 | + Given [5, 7, 7, 8, 8, 10] and target value 8, |
| 16 | + return [3, 4]. |
| 17 | +
|
| 18 | + Solution: It takes O(lgN) to find both the lower-bound and upper-bound. |
| 19 | + */ |
| 20 | +publicclassSolution { |
| 21 | +publicint[]searchRange(int[]A,inttarget) { |
| 22 | +int[]res =newint[]{-1,-1}; |
| 23 | +intlower =getLowerBound(A,target); |
| 24 | +intupper =getUpperBound(A,target); |
| 25 | +if (lower <=upper) { |
| 26 | +res[0] =lower; |
| 27 | +res[1] =upper; |
| 28 | + } |
| 29 | +returnres; |
| 30 | + } |
| 31 | +publicintgetLowerBound(int[]A,inttarget) { |
| 32 | +intl =0,r =A.length -1; |
| 33 | +while (l <=r) { |
| 34 | +intmid = (l+r) /2; |
| 35 | +if (A[mid] <target)l =mid +1; |
| 36 | +elser =mid -1; |
| 37 | + } |
| 38 | +returnl; |
| 39 | + } |
| 40 | +publicintgetUpperBound(int[]A,inttarget) { |
| 41 | +intl =0,r =A.length -1; |
| 42 | +while (l <=r) { |
| 43 | +intmid = (l+r) /2; |
| 44 | +if (A[mid] <=target)l =mid +1; |
| 45 | +elser =mid -1; |
| 46 | + } |
| 47 | +returnr; |
| 48 | + } |
| 49 | +} |
