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]
class Solution { public int[] searchRange(int[] nums, int target) { int l = 0, r = nums.length - 1; if (nums.length == 0) return new int[]{-1, -1}; while (l < r) { int mid = (l + r) >> 1; if (nums[mid] >= target) { r = mid; } else { l = mid + 1; } } int []ans = new int[2]; if (nums[l] == target) ans[0] = l; else ans[0] = -1; System.out.println(ans[0]); l = 0; r = nums.length - 1; while (l < r) { int mid = (l + r) >> 1; if (nums[mid] <= target) { l = mid + 1; } else { r = mid - 1; } } if (r < nums.length && r >= 0 && nums[r] == target) { ans[1] = r; } else if (r - 1 >= 0 && nums[r - 1] == target) { ans[1] = r - 1; } else ans[1] = -1; return ans; } }
34. Find First and Last Position of Element in Sorted Array
原文：https://www.cnblogs.com/hyxsolitude/p/12310151.html