2014.2.13 20:27
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
Solution1:
Apparently this problem is a test for binary search. You could either use the function upper_bound() and lower_bound() provided by STL <algorithm>, or write one for yourself.
Total time complexity is O(log(n)). Space complexity is O(1).
Accepted code:
1 // 1AC, the code can be shorter, but..1AC is just fine. 2 class Solution { 3 public: 4 vector<int> searchRange(int A[], int n, int target) { 5 vector<int> res; 6 int ll, rr, mm; 7 int i1, i2; 8 9 if (A == nullptr || n <= 0) { 10 res.push_back(-1); 11 res.push_back(-1); 12 return res; 13 } 14 15 if (n == 1) { 16 if (A[0] == target) { 17 res.push_back(0); 18 res.push_back(0); 19 } else { 20 res.push_back(-1); 21 res.push_back(-1); 22 } 23 return res; 24 } 25 26 if (target < A[0] || target > A[n - 1]) { 27 res.push_back(-1); 28 res.push_back(-1); 29 return res; 30 } 31 32 if (target == A[0]) { 33 i1 = 0; 34 } else { 35 ll = 0; 36 rr = n - 1; 37 while (rr - ll > 1) { 38 mm = (ll + rr) / 2; 39 if (A[mm] < target) { 40 ll = mm; 41 } else { 42 rr = mm; 43 } 44 } 45 if (A[rr] != target) { 46 res.push_back(-1); 47 res.push_back(-1); 48 return res; 49 } else { 50 i1 = rr; 51 } 52 } 53 54 if (target == A[n - 1]) { 55 i2 = n - 1; 56 } else { 57 ll = 0; 58 rr = n - 1; 59 while (rr - ll > 1) { 60 mm = (ll + rr) / 2; 61 if (A[mm] > target) { 62 rr = mm; 63 } else { 64 ll = mm; 65 } 66 } 67 if (A[ll] != target) { 68 res.push_back(-1); 69 res.push_back(-1); 70 return res; 71 } else { 72 i2 = ll; 73 } 74 } 75 res.push_back(i1); 76 res.push_back(i2); 77 return res; 78 } 79 };
Solution2:
The STL version.
Accepted code:
1 // 1AC, very smooth. 2 #include <algorithm> 3 using namespace std; 4 5 class Solution { 6 public: 7 vector<int> searchRange(int A[], int n, int target) { 8 vector<int> res; 9 10 res.push_back(-1); 11 res.push_back(-1); 12 if (A == nullptr) { 13 return res; 14 } 15 16 int *px, *py; 17 int ix, iy; 18 19 px = lower_bound(A, A + n, target); 20 ix = px - A; 21 if (ix >= n || A[ix] != target) { 22 // not found 23 return res; 24 } 25 26 py = upper_bound(A, A + n, target); 27 --py; 28 iy = py - A; 29 res[0] = ix; 30 res[1] = iy; 31 32 return res; 33 } 34 };