34. Search for a Range

Posted 2020-11-01 Zzz...y

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].

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

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

找目标值涵盖的区间，要求时间复杂度O(logn)。

看到O(logn)，直接想到的就是二分查找。然后想到用两次二分法，分别求第一个下界和上界。

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