33. Search in Rotated Sorted Array

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

 

想法：发现此旋转数组在某个数字前后是顺序排列的，所以我们采用二分查找的方法。

其思想是，当旋转数组时，必须有一半数组仍按排序顺序排列。              例如，6 7 1 2 3 4 5，顺序从7到1之间的点中断。所以在进行二进制搜索时，我们可以判断哪个部分是有序的，目标是否在这个范围内，如果是的话，继续在这个范围内搜索，如果不是，继续在另一个范围内搜索。

注意：start可能=end  target有可能等于 num[start]获nums[end] 要考虑到全体的情况 还有查找的大条件是while循环。

class Solution {
    public int search(int[] nums, int target) {
        int start =0;
        int end = nums.length-1;
        
      while(start <= end){
            int mid = (start+end)/2;
            if(nums[mid]==target){
                return mid;
            }
            if(nums[start]<=nums[mid]){
                if(nums[start]<=target && nums[mid]>target)
                    end = mid-1;
                else
                    start=mid+1;
            }
            if(nums[mid]<=nums[end]){
                if(nums[mid]<target && nums[end]>=target)
                    start = mid+1;
                else
                    end=mid-1;
            }
            
      }
        return -1;
    }
}