leetcode -- Search for a Range (TODO)

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].

思路：

使用二分搜索找到target的idx，然后查看该idx的左右确定范围。

算法复杂度：

平均情况下是O(lgn);

最坏情况下数组中所有元素都相同O(n);

 public class Solution {
     public int[] searchRange(int[] A, int target) {
         // Start typing your Java solution below
         // DO NOT write main() function
         int idx = binarySearch(A, target);
         int len = A.length;
         int[] results = null;
         if(idx == -1){
             results = new int[]{-1, -1};
         } else{
             int l = idx;
             int r = idx;
             while(l >= 0 && A[l] == target){
                 l--;
             }
             l++;

             while(r < len && A[r] == target){
                 r++;
             }
             r--;
             results = new int[]{l, r};
         }
         return results;
     }

     public int binarySearch(int[] A, int target){
         int len = A.length;
         int l = 0, r = len - 1;
         while(l <= r){
             int mid = (l + r) / 2;
             if(target == A[mid])
                 return mid;

             if(target > A[mid]){
                 l = mid + 1;
             } else {
                 r = mid - 1;
             }
         }

         return -1;
     }
 }

google了下，要保证最坏情况下时间复杂度为O(lgn)：进行两次二分搜索确定左右边界