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].

算法_1

stl-使用lower_bound和upper_bound

算法_2

直接二分查找修改

复杂度

O(lgn)

#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;

}

}
};

#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;

}

};