在排序数组中查找元素的第一个和最后一个位置

原创已于 2024-03-05 16:43:29 修改 · 284 阅读

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#算法 #leetcode #数据结构

于 2024-03-05 16:43:10 首次发布

该文章介绍了一个在已排序整数数组中查找目标值及其出现范围的算法，利用二分查找法优化时间复杂度至O(logn)，并提供了示例和代码实现。

在排序数组中查找元素的第一个和最后一个位置

给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

进阶：你可以设计并实现时间复杂度为 $O(log⁡n)O(\log n)$ 的算法解决此问题吗？

示例 1：

输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]
示例 2：

输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]
示例 3：

输入：nums = [], target = 0
输出：[-1,-1]

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> results;
        int left = 0, right = nums.size()-1;
        int mid = (left + right)/2;
        int result1 = -1, result2 = -1, result = -1;
        int target1 = target - 1, target2 = target + 1;
        while(left<=right){
            if(nums[mid]==target){
                result = mid;
                break;
            }else if(nums[mid]>target1){
                right = mid -1 ;
                mid = (left + right)/2;
            }else{
                left = mid + 1;
                mid = (left + right)/2;
            }
        }
        if(result==-1){
            results.push_back(-1);
            results.push_back(-1);
            return results;
        }
        left = 0;
        right = nums.size()-1;
        mid = (left + right)/2;
        while(left<=right){
            if(nums[mid]==target1){
                result1 = mid;
                break;
            }else if(nums[mid]>target1){
                right = mid -1 ;
                mid = (left + right)/2;
            }else{
                left = mid + 1;
                mid = (left + right)/2;
            }
        }
        if(result1==-1){
            if(mid==0&&nums[0]>target1){
                result1 = -1;
            }else{
                result1 = mid;
            } 
        }else if(result1!=nums.size()-1){
            while(nums[result1]==nums[result1+1]&&result1<nums.size()+1)
            {
                result1++;
            }
        }
        left = 0;
        right = nums.size()-1;
        mid = (left + right)/2;
        while(left<=right){
            if(nums[mid]==target2){
                result2 = mid;
                break;
            }else if(nums[mid]>target2){
                right = mid -1 ;
                mid = (left + right)/2;
            }else{
                left = mid + 1;
                mid = (left + right)/2;
            }
        }
        if(result2==-1){
            if(mid==0&&nums[0]>target2){
                result2 = -1;
            }else{
                result2 = mid+1;
            } 
        }else if(result2!=0){
            while(nums[result2]==nums[result2-1]&&result2>0)
            {
                result2--;
            }
        }
        if(result1+1>result2-1){
            results.push_back(result2-1);
            results.push_back(result2-1);
        }else{
            results.push_back(result1+1);
            results.push_back(result2-1);  
        }     
        return results;
    }
};