33. Search in Rotated Sorted Array

题目

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

思路

本题数据“亚有序”本来可以用二分查找的，思路就是先用二分法找最小值，时间复杂度log(n)，再从最小值及数组首端处或最小值处和数组末端处用二分查找。
但是，比较懒，就直接用了标准库函数

代码

class Solution {public:    int search(vector<int>& nums, int target) {        int index = find(nums.begin(),nums.end(),target)-nums.begin();        return index>=0&&index<nums.size()?index:-1;    }};