UR機械臂的TCP/IP控制
阿新 • • 發佈:2021-11-14
題目連結:https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
題目描述:
給定一個按照升序排列的整數陣列 nums,和一個目標值 target。找出給定目標值在陣列中的開始位置和結束位置。
如果陣列中不存在目標值 target,返回[-1, -1]。
進階:
你可以設計並實現時間複雜度為O(log n)的演算法解決此問題嗎?
示例 1:
輸入:nums = [5,7,7,8,8,10], target = 8
輸出:[3,4]
示例2:
輸入:nums = [5,7,7,8,8,10], target = 6
輸出:[-1,-1]
示例 3:
輸入:nums = [], target = 0
輸出:[-1,-1]
提示:
0 <= nums.length <= 105
-109<= nums[i]<= 109
nums是一個非遞減陣列
-109<= target<= 109
題解:
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { int leftBoard = getLeftSearch(nums, target); int rightBoard = getRightSearch(nums, target); if(leftBoard <= rightBoard && nums[leftBoard] == target && nums[rightBoard] == target) return {leftBoard, rightBoard}; return {-1, -1}; } int getLeftSearch(vector<int>& nums, int target) { int left = -1; int right = nums.size(); while(left + 1 != right) { int mid = left + (right - left) / 2; if(nums[mid] < target) left = mid; else right = mid; } return right; } int getRightSearch(vector<int>& nums, int target) { int left = -1; int right = nums.size(); while(left + 1 != right) { int mid = left + (right - left) / 2; if(nums[mid] <= target) left = mid; else right = mid; } return left; } };