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给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。

如果目标值不在数组中，则返回[-1, -1]

样例

给出[5, 7, 7, 8, 8, 10]和目标值target=8,

返回[3, 4]

挑战

时间复杂度 O(log n)

标签

数组 排序数组 二分法

 1 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:
 8     vector<int> searchRange(vector<int> &A, int target) {
 9         // write your code here
10         vector<int> result;
11         int size = A.size()-1;
12         if(A.empty()) {
13             result.push_back(-1);
14             result.push_back(-1);
15         }
16         else {
17             result.push_back(getFirstTarget(A, target, 0, size));
18             result.push_back(getLastTarget(A, target, 0, size));
19         }
20         return result;
21     }
22
23     int getFirstTarget(vector<int> &A, int target, int low, int high) {
24         if(low > high)
25             return -1;
26         int mid = (low+high)/2;
27         if(A[mid] == target) {
28             if((mid>0 && A[mid-1]!=target) || mid==0)
29                 return mid;
30             else
31                 high = mid-1;
32         }
33         else if(A[mid] > target)
34             high = mid-1;
35         else
36             low = mid+1;
37
38         return getFirstTarget(A, target, low, high);
39     }
40
41     int getLastTarget(vector<int> &A, int target, int low, int high) {
42         int size = A.size()-1;
43         if(low > high)
44             return -1;
45         int mid = (low+high)/2;
46         if(A[mid] == target)  {
47             if((mid<size && A[mid+1]!=target) || mid==size)
48                 return mid;
49             else
50                 low = mid+1;
51         }
52         else if(A[mid] > target)
53             high = mid-1;
54         else
55             low = mid+1;
56
57         return getLastTarget(A, target, low, high);
58     }
59 };