LeetCode 34. Search for a Range

寻找目标区间
本文介绍了一种在有序整数数组中查找指定目标值起始和结束位置的算法,该算法的时间复杂度为O(log n),并提供了C++实现示例。

Given an array of integers 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> v(2);
        int l=0,r=nums.size()-1,m=0;
        while(l<=r){
            m=l+(r-l)/2;
            if(nums[m]>target) r=m-1;
            else if(nums[m]<target) l=m+1;
            else break;
        }
        if(l>r){
            v[0]=-1;v[1]=-1;return v;
        }
        while(nums[l]!=target) l++;
        while(nums[r]!=target) r--;
        v[0]=l;v[1]=r;
        return v;
    }
};

### LeetCode Problem 34: Find First and Last Position of Element in Sorted Array The task involves finding the starting and ending position of a given target value within an array of integers. If the target is not found in the array, [-1, -1] should be returned. For instance, consider an input where `nums` = [5,7,7,8,8,10], and `target` = 8. The expected output would be [3, 4]. Another example could involve `nums` = [5,7,7,8,8,10], but this time with `target` = 6, leading to an output of [-1, -1]. To solve this problem efficiently: A binary search approach can achieve logarithmic complexity by narrowing down potential positions for both the first and last occurrence of the target element[^1]: ```python def searchRange(nums, target): def findLeftIndex(nums, target): left, right = 0, len(nums) - 1 while left <= right: mid = (left + right) // 2 if nums[mid] < target: left = mid + 1 else: right = mid - 1 return left def findRightIndex(nums, target): left, right = 0, len(nums) - 1 while left <= right: mid = (left + right) // 2 if nums[mid] <= target: left = mid + 1 else: right = mid - 1 return right left_index = findLeftIndex(nums, target) right_index = findRightIndex(nums, target) # Check if the target exists in the list. if left_index <= right_index < len(nums) and nums[left_index] == target: return [left_index, right_index] return [-1, -1] ``` This code snippet defines two helper functions that perform modified versions of binary searches—one looking for the start index (`findLeftIndex`) and another for the end index (`findRightIndex`). After determining these indices, it checks whether they are valid before returning them as part of the result.
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值