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.

Binary Search

Time Complexity

O(logn)Space ComplexityO(1)

思路

If target == nums[mid]; return mid;

If target nums[mid] < nums[right], mid to right is sorted, so we just need to know if target is between mid to right. If target nums[mid] > nums[right], left to mid is sorted, so we just need to know if target is between left to mid.

代码

public int search(int[] nums, int target) {    if(nums == null || nums.length == 0){        return -1;    }    int start = 0, end = nums.length - 1;    while(start + 1 < end){        int mid = start + (end - start)/2;        if(nums[mid] == target){            return mid;        }        if(nums[mid] < nums[end]){    //right part is sorted            if(nums[mid] < target && target <= nums[end]){                start = mid + 1;            }else{                end = mid - 1;            }        }else if(nums[mid] > nums[end]){   //left part is sorted            if(nums[start] <= target && target < nums[mid]){                end = mid - 1;            }else{                start = mid + 1;            }        }else{            start++;        }    }    if(nums[start] == target){        return start;    }    if(nums[end] == target){        return end;    }    return -1;}