[leetcode]34.Search for a Range找区间

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]

题意：

给定一个有序数组，找出某个值的起始和终止区间。

思路：

二分查找

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代码：

 1 class Solution {
 2        public int[] searchRange(int[] nums, int target) {
 3         if(nums == null || nums.length == 0) return new int[]{-1,-1};
 4 
 5         int left = lower_bound(nums, 0, nums.length-1, target);
 6         if(left == -1) return new int[]{-1,-1};
 7         int right = upper_bound(nums, 0, nums.length-1, target);
 8         return new int[]{left, right};
 9 
10     }
11 
12     int lower_bound (int[] nums, int left, int right, int target) {
13         while(left + 1 < right){
14             int mid =  left + (right - left)/2;
15             if(nums[mid] < target){
16                 left = mid;
17             }else{
18                 right = mid;
19             }
20         }
21         if (nums[left] == target) return left;
22         if(nums[right] == target) return right;
23         return -1;
24     }
25 
26     int upper_bound (int[] nums, int left, int right, int target) {
27         while(left + 1 < right){
28             int mid =  left + (right - left)/2;
29             if(nums[mid] > target){
30                 right = mid;
31             }else{
32                 left = mid;
33             }
34         }
35         if(nums[right] == target) return right;
36         if (nums[left] == target) return left;
37         return -1;
38     }
39 }