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

class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) return -1;
        int l = 0, r = nums.length - 1;
        int n = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (target < nums[mid]) {
                if (nums[r] > nums[l] || nums[mid] <= nums[r]) r = mid - 1;
                else {
                    if (target <= nums[r]) l = mid + 1;
                    else r = mid - 1;
                }
            } else if (target > nums[mid]) {
                if (nums[r] > nums[l] || nums[mid] >= nums[l]) l = mid + 1;
                else {
                    if (target <= nums[r]) {
                        l = mid + 1;
                    } else {
                        r = mid - 1;
                    }
                }
            } else {
                return mid;
            }
        }
        if (l >= n || r < 0) return -1;
        return nums[l] == target ? l: -1;
    }
}

class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) return -1;
        int n = nums.length;
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] > nums[r]) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        System.out.println(l);
        int offset = l;
        l = 0;
        r = n - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            //System.out.println(l + " " + mid + " " + r);
            if (target < nums[(mid + offset) % n]) {
                r = mid - 1;
            } else if(target > nums[(mid + offset) % n]) {
                l = mid + 1;
                
            } else {
                return (mid + offset) % n;
            }
        }
        
        return nums[(l + offset) % n] == target ? (l + offset) % n : -1;
        
    }
}

class Solution {
    public int check(int[] nums, int target, int s, int e) {
        
        
        int l = s, r = e;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (target < nums[mid]) {
                r = mid - 1;
            } else if (target > nums[mid]) {
                l = mid + 1;
                
            } else {
                return mid;
            }
        }
        
        System.out.println("l = " + l);
        return nums[l] == target ? l : -1;
        
    }
    public int search(int[] nums, int target) {
        int n = nums.length;
        if (n == 0) return -1;
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[l] < nums[mid]) {
                int temp = check(nums, target, l, mid);
                if (temp != -1) {
                    return temp;
                } else {
                    l = mid + 1;
                }
            } else {
                int temp = check(nums, target, mid + 1, r);
                if (temp != -1) {
                    return temp;
                } else {
                    r = mid;
                }
            }
        }
        return nums[l] == target ? l : -1;
        
    }
}