leetcode 34. Find First and Last Position of Element in Sorted Array

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]

 

思路：二分

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         vector<int> v(2, -1);
 5         int l = 0, r = nums.size() - 1;
 6         while (l < r) {
 7             int mid = ((r - l) >> 1) + l;
 8             if (nums[mid] >= target) //尽可能向左靠
 9                 r = mid;
10             else
11                 l = mid + 1;
12         }
13         if (r < 0 || nums[l] != target)
14             return v;
15         v[0] = l;
16         r = nums.size() - 1;
17         while (l < r) {
18             int mid = ((r - l) >> 1) + l + 1; //偏向于右边
19             if (nums[mid] <= target) //尽可能向右靠
20                 l = mid;
21             else
22                 r = mid - 1;
23         }
24         v[1] = l;
25         return v;
26     }
27 };

 

posted @ 2019-09-16 23:22  琴影  阅读(161)  评论(0)  编辑  收藏  举报