leetcode_c++：Search for a Range（034）

题目

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].

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算法_1

1. stl-使用lower_bound和upper_bound

算法_2

1. 直接二分查找修改

复杂度

O(lgn)

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#include<iostream>#include<vector>#include <algorithm>using namespace std;class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int>::iterator lower=lower_bound(nums.begin(), nums.end(), target);        //lower_bound返回第一个大于等于value值得位置        vector<int>::iterator upper=upper_bound(nums.begin(), nums.end(), target);        //upper 是第一个大于value值得位置        vector<int> resultNo;        resultNo.push_back(-1);        resultNo.push_back(-1);        vector<int> ret;        ret.push_back(lower-nums.begin());        ret.push_back(upper-nums.begin()-1);        if(*lower !=target){            return resultNo;        }else{            return ret;        }    }};

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#include<iostream>#include<vector>#include <algorithm>using namespace std;class Solution {public:    vector<int> searchRange(vector<int>& nums, int target){        vector<int> ret;        ret.push_back(-1);        ret.push_back(-1);        int left=0,right=nums.size()-1,mid;        while(left<=right){            if(nums[left]==target && nums[right]==target){                ret[0]=left;                ret[1]=right;                break;            }            mid=left+(right-left)/2;            if(nums[mid]<target){                left=mid+1;            }else if(nums[mid]>target){                right=mid-1;            }else{                if(nums[right]==target)                    ++left;                else                    --right;            }        }    return ret;    }};