Given an array of integers nums
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]
.
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1]
解题思路:二分法先找到一个target
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int low=0,high=nums.size()-1,result=-1;
vector<int>a;
while(low<=high){
int mid=(low+high)/2;
if(nums[mid]==target){
result=mid;
break;
}
else if(nums[mid]>target)
high=mid-1;
else
low=mid+1;
}
if(result==-1){
a.push_back(-1);
a.push_back(-1);
}
else{
int p=result,q=result;
while(p>=0&&nums[p]==target)
p--;
while(q<nums.size()&&nums[q]==target)
q++;
a.push_back(p+1);
a.push_back(q-1);
}
return a;
}
};