34. Find First and Last Position of Element in Sorted Array-优快云博客

本文链接：https://blog.youkuaiyun.com/csg3140100993/article/details/102822907

本文介绍了一种算法，用于在升序排列的整数数组中找到特定目标值的开始和结束位置。算法的时间复杂度为O(log n)，满足高效查找的需求。若目标值不存在于数组中，则返回[-1,-1]。

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

Medium

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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]

Accepted

369,381

Submissions

1,071,852

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> rt;
        rt.push_back(searchlow(nums,target));
        rt.push_back(searchhigh(nums,target));
        return rt;
    }
    int searchlow(vector<int>& nums, int target){
    	int left=0;
    	int right=nums.size()-1;
    	while(left<=right){
    		int mid=(left+right)>>1;
    		
    		if(nums[mid]==target){
    			while(mid>=0&&nums[mid]==target){
    				mid--;
    			}
    			mid++;
    			return mid;
    		}
    		if(nums[mid]<target){
    			left=mid+1;
    		}else{
    			right=mid-1;
    		}
    	}
    	return -1;
    }
    int searchhigh(vector<int>& nums, int target){
    	int left=0;
    	int right=nums.size()-1;
    	while(left<=right){
    		int mid=(left+right)>>1;
    		
    		if(nums[mid]==target){
    			while(mid<=(nums.size()-1)&&nums[mid]==target){
    				mid++;
    			}
    			mid++;
    			return mid;
    		}
    		if(nums[mid]<target){
    			left=mid+1;
    		}else{
    			right=mid-1;
    		}
    	}
    	return -1;
    }
};