34. Search for a Range

题目

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].

翻译

给定按升序排序的整数数组，找到给定目标值的起始和终止位置。

您的算法的运行时复杂度必须是O（log n）的顺序。

如果在数组中找不到目标，返回[-1, -1]。

例如，
给定[5, 7, 7, 8, 8, 10]和目标值8，
返回[3, 4]。

分析

采用二分法，因为要求target的左极限与右极限，所以要用二分法查找两次

时间复杂度O(logn)

递归法：

class Solution {
public:
    int Binaryup(vector<int>& nums,int target,int begin,int end){
        if(begin>end) return end;
        int mid=begin+(end-begin)/2;
        if(nums[mid]>target)  return Binaryup(nums,target,begin,mid-1);
        else return Binaryup(nums,target,mid+1,end);
    }
    
    int Binarylow(vector<int>& nums,int target,int begin,int end){
        if(begin>end)  return begin;
        int mid=begin+(end-begin)/2;
        if(nums[mid]<target) return Binarylow(nums,target,mid+1,end);
        else return Binarylow(nums,target,begin,mid-1);
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result(2,-1);
        int h=Binaryup(nums,target,0,nums.size()-1);
        int l=Binarylow(nums,target,0,nums.size()-1);
        if(h>=l){
            result[0]=l;
            result[1]=h;
            return result;
        }
        return result;
    }
};

while循环：

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