Search for a Range 61
Question
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.
Solution
做两遍binary search。第一遍找第一个出现的target,第二遍找最后出现的target。
代码如下:
public class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public int[] searchRange(int[] A, int target) {
// write your code here
if (A == null || A.length == 0){
return new int[]{-1, -1};
}
int start = 0;
int end = A.length - 1;
int[] bound = new int[2];
while (start + 1 < end){
int mid = start + (end - start)/2;
if (A[mid] == target){
end = mid;
}else if (A[mid] < target){
start = mid;
}else{
end = mid;
}
}
if (A[start] == target){
bound[0] = start;
}else if (A[end] == target){
bound[0] = end;
}else{
bound[0]=bound[1]=-1;
return bound;
}
start = 0; end = A.length - 1;
while (start + 1 < end){
int mid = start + (end - start)/2;
if (A[mid] == target){
start = mid;
}else if (A[mid] < target){
start = mid;
}else{
end = mid;
}
}
if (A[end] == target){
bound[1] = end;
}else if (A[start] == target){
bound[1] = start;
}else{
bound[0]=bound[1]=-1;
return bound;
}
return bound;
}
}