Given an array of integers sorted in ascending order, 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].
給定按升序排序的整數數組，找到給定目標值的開始和結束位置。算法的時間複雜度爲O（log n）。如果在數組中找不到目標，則返回[-1，-1]。
代碼如下:
public class SearchforARange {
public static void main(String[] args) {
int[] nums = new int[]{5, 7, 7, 8, 8, 10};
int[] searchRange = searchRange(nums, 8);
for (int i = 0; i < 2; i++) {
System.out.print(searchRange[i]+",");
}
}
public static int[] searchRange(int[] nums, int target) {
int[] result = new int[]{-1,-1};
if(nums.length < 0) {
return result;
}
int start=0;
int end = nums.length-1;
for(;start<nums.length;start++) {
if(nums[start] == target) {
result[0]=start;
break;
}
}
for(;end>=start;end--){
if(nums[end] == target) {
result[1]=end;
break;
}
}
return result;
}
}