[LintCode] Search for a Range

时间：2017-09-08 13:38:35 阅读：161 评论：0 收藏：0 [点我收藏+]

标签：art query starting ast first body logs [1] else

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].

Challenge

O(log n) time.

Break this problem into two subproblems:

1. find the first element in the array of the given target value;

2. find the last element in the array of the given target value;

Both of these two subproblems can be solved by twisting the regular binary search algorithm when the target value is found.

To find the first occurence, keep looking its left range, including the mid element;

To find the last occurence, keep looking its right range, including the mid element;

 1 public class Solution {
 2     /** 
 3      *@param A : an integer sorted array
 4      *@param target :  an integer to be inserted
 5      *return : a list of length 2, [index1, index2]
 6      */
 7     public int[] searchRange(int[] A, int target) {
 8         int[] result = new int[2];
 9         result[0] = -1;
10         result[1] = -1;
11         if(A == null || A.length == 0)
12         {
13             return result;
14         }
15         result[0] = searchRangeLeft(A, target, 0, A.length - 1);
16         if(result[0] < 0)
17         {
18             return result;
19         }
20         else
21         {
22             result[1] = searchRangeRight(A, target, 0, A.length - 1);
23         }
24         return result;
25     }
26     private int searchRangeLeft(int[] A, int target, int lo, int hi)
27     {
28         if(hi - lo <= 1)
29         {
30             if(A[lo] == target)
31             {
32                 return lo;
33             }
34             else if(A[hi] == target)
35             {
36                 return hi;
37             }
38             else
39             {
40                 return -1;
41             }
42         }
43         int mid = lo + (hi - lo) / 2;
44         if(A[mid] == target)
45         {
46             return searchRangeLeft(A, target, lo, mid);
47         }
48         else if(A[mid] > target)
49         {
50             return searchRangeLeft(A, target, lo, mid - 1);
51         }
52         else
53         {
54             return searchRangeLeft(A, target, mid + 1, hi);
55         }
56     }
57     private int searchRangeRight(int[] A, int target, int lo, int hi)
58     {
59         if(hi - lo <= 1)
60         {
61             if(A[hi] == target)
62             {
63                 return hi;
64             }
65             else if(A[lo] == target)
66             {
67                 return lo;
68             }
69             else
70             {
71                 return -1;
72             }
73         }
74         int mid = lo + (hi - lo) / 2;
75         if(A[mid] == target)
76         {
77             return searchRangeRight(A, target, mid, hi);
78         }
79         else if(A[mid] > target)
80         {
81             return searchRangeRight(A, target, lo, mid - 1);
82         }
83         else
84         {
85             return searchRangeRight(A, target, mid + 1, hi);
86         }
87     }
88 }