Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
 

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

題意

找已排序數組中某個值的範圍

題解

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         int size = nums.size(), s = 0, e = size - 1;
 5         vector<int>ans;
 
 6         if (size <= 0||target<nums[0]||target>nums[size-1]) {
 7             ans.push_back(-1);
 8             ans.push_back(-1);
 9             return ans;
10         }
11         while (s < e) {
12             int mid = (s + e) / 2;
13             if (nums[mid] < target)
14                 s = mid + 1
 
;
15             else
16                 e = mid;
17         }
18         if (nums[s] != target) {
19             ans.push_back(-1);
20             ans.push_back(-1);
21             return ans;
22         }
23         ans.push_back(s);
24         s = 0, e = size - 1;
25         while (s <= e) {
26             int mid = (s + e) / 2;
27             if (nums[mid] <= target)
28                 s = mid + 1;
29             else
30                 e = mid - 1;
31         }
32         ans.push_back(e);
33         return ans;
34     }
35 };

兩次二分

19.2.2 [LeetCode 34] Find First and Last Position of Element in Sorted Array