LeetCode之Search for a Range_hoieporm-优快云博客

本文介绍了一种算法，可在已排序的整数数组中找到特定目标值的起始和结束位置，该算法的时间复杂度为O(log n)。如果未找到目标值，则返回[-1,-1]。示例：给定数组[5,7,7,8,8,10]及目标值8，返回[3,4]。

Given a sorted array of integers, 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(int A[], int n, int target) {
        vector<int> ivec;
        int left=lower_bound(A,n,target);
        if(left==n||A[left]!=target) {
            ivec.push_back(-1);
            ivec.push_back(-1);
            return ivec;
        } 
        ivec.push_back(left);
        int right=upper_bound(A,n,target);
        ivec.push_back(right-1);
        return ivec;
        
    }
    //可插入target的第一个合适位置
    int lower_bound(int A[],int n,int target){
        int first=0,len=n,half;
        int mid;
        while(len>0){
            half=len>>1;
            mid=first+half;
            if(A[mid]<target){
                first=mid+1;
                len=len-half-1;
            }
            else
                len=half;
        }
        return first;
    }
    //可插入target的最后一个合适位置
    int upper_bound(int A[],int n,int target){
        int first=0,len=n,half;
        int mid;
        while(len>0){
            half=len>>1;
            mid=first+half;
            if(A[mid]>target){
                len=half;
            }
            else{
               first=mid+1;
                len=len-half-1; 
            }
        }
        return first;
    }
};