2019.2.12
题目描述:
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]
这题是查找给定的target在一个升序数组中的首尾index,不过要求时间复杂度为O(log n)。
解法一:
因为是升序数组且要求时间复杂度是O(log n),所以可以考虑二分法查找。利用两次二分查找,先查找左边界,再查找右边界即可。
C++代码:
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res(2,-1);
if(nums.size()==0) return res;
int low=0,high=nums.size()-1;
while(low<high){
int mid=low+(high-low)/2;
if(nums[mid]<target)
low=mid+1;
else
high=mid;
}
if(nums[high]!=target)
return res;
res[0]=high;
high=nums.size();
while(low<high){
int mid=low+(high-low)/2;
if(nums[mid]<=target)
low=mid+1;
else
high=mid;
}
res[1]=low-1;
return res;
}
};