【LEET-CODE】33. Search in Rotated Sorted Array【Medium】

Question:

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：

给定一个升序数组，不过首位不一定是最小值，因为数组进行了若干次循环移动，求target是否在数组中，如果在则返回索引，不在返回-1。

代码很简单，最多循环数组size次退出循环，逐位比较，直到找到target或者数组不可能存在target时，结束循环。

Code:

class Solution {public:    int search(vector<int>& nums, int target) {        int j = 0, k;        for(int i = 0; i < nums.size() ; i++){            k=j;            if(nums[j]==target) return j;            else if(nums[j]> target) j--;            else if(nums[j]< target) j++;            if(j<0) j=nums.size()-1;            else if(j>nums.size()-1) j=0;            if( (nums[j]>target && nums[k]<target)||(nums[k]>target && nums[j]<target) )            //如果target在两个相邻的数中间，则返回-1            return -1;        }        return -1;    }};