Search for a Range

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

思路：二分查找。使用两次二分查找。第一次找到第一个等于target的数组下标，第二次找到最后一个等于target的数组的下标。

 1 public class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         if (A.length == 0) {
 4             return new int[]{-1, -1};
 5         }
 6         
 7         int start, end, mid;
 8         int[] bound = new int[2]; 
 9         
10         // search for left bound
11         start = 0; 
12         end = A.length - 1;
13         while (start + 1 < end) {
14             mid = start + (end - start) / 2;
15             if (A[mid] == target) {
16                 end = mid;
17             } else if (A[mid] < target) {
18                 start = mid;
19             } else {
20                 end = mid;
21             }
22         }
23         if (A[start] == target) {
24             bound[0] = start;
25         } else if (A[end] == target) {
26             bound[0] = end;
27         } else {
28             bound[0] = bound[1] = -1;
29             return bound;
30         }
31         
32         // search for right bound
33         start = 0;
34         end = A.length - 1;
35         while (start + 1 < end) {
36             mid = start + (end - start) / 2;
37             if (A[mid] == target) {
38                 start = mid;
39             } else if (A[mid] < target) {
40                 start = mid;
41             } else {
42                 end = mid;
43             }
44         }
45         if (A[end] == target) {
46             bound[1] = end;
47         } else if (A[start] == target) {
48             bound[1] = start;
49         } else {
50             bound[0] = bound[1] = -1;
51             return bound;
52         }
53         
54         return bound;
55     }
56 }

 

转载于:https://www.cnblogs.com/FLAGyuri/p/5370647.html