题目
思路
- 用find函数找其位置下标,最后返回
这样find()的时间复杂度是O(n)(实际上是遍历)
后来看到官网题解,可以用二分法,时间复杂度为O(logn)
代码
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int search1(vector<int>& nums, int target) {
auto pos = find(nums.begin(),nums.end(),target);
if(pos!=nums.end()) return pos-nums.begin();
else return -1;
}
int search(vector<int>& nums, int target) {
//官网题解
int size = nums.size();
int left = 0, right = size-1;
while(left<=right){
int mid = (left + right)>>1;
if(nums[mid] == target) return mid;
if(nums[left] <= nums[mid]) {
if(nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if(nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}
int main(){
vector<int> nums;
nums.emplace_back(4);
nums.emplace_back(5);
nums.emplace_back(6);
nums.emplace_back(7);
nums.emplace_back(0);
nums.emplace_back(1);
nums.emplace_back(2);
int pos = search(nums, 0);
cout<<pos<<endl;
return 0;
}