[LintCode] 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]
.
Given [5, 7, 7, 8, 8, 10]
and target value 8
,
return [3, 4]
.
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 }
Related Problems
Total Occurrence of Target
Range Sum Query 2D - Immutable