33. Search in Rotated Sorted Array (JAVA)

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place and use only constant extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

O(log n) =>想到二分法

class Solution {
    public int search(int[] nums, int target) {
        return binarySearch(nums,target,0,nums.length-1); 

    }
    public int binarySearch(int[] nums, int target, int left, int right){
        if(left > right) return -1; //not found

        int mid = left + ((right-left)>>1);
        if(target == nums[mid]) return mid; //found

        if(nums[mid] < nums[left]){//rotate in the left part
            if(target < nums[mid] || target >nums[right]){ //target in the left part
                return binarySearch(nums,target,left,mid-1);
            }
            else{ //target in the right part
                return binarySearch(nums,target,mid+1,right);
            }
        }
        else if(nums[mid] > nums[right]){ //rotate in the right part
            if(target > nums[mid] || target < nums[left]) {//target in the right part
                return binarySearch(nums,target,mid+1,right);
            }
            else{//target in the left part
                return binarySearch(nums,target,left,mid-1);
            }
        }
        else{ //no rotate
            if(target < nums[mid]){
                return binarySearch(nums,target,left,mid-1);
            }
            else{
                return binarySearch(nums,target,mid+1,right);
            }
        }

    }
}