用 Java 构建旋转排序数组搜索：了解枢轴搜索和二分搜索

[1, 2, 3, 4, 5, 6]

[4, 5, 6, 1, 2, 3]

class Solution {
    public static void main(String[] args) {
        int[] arr = {4, 5, 6, 1, 2, 3}; // Example of rotated sorted array
        int target = 5;

        // Searching for the target
        int result = search(arr, target);

        // Displaying the result
        System.out.println("Index of target: " + result);
    }

    // Main search function to find the target in a rotated sorted array
    public static int search(int[] nums, int target) {
        // Step 1: Find the pivot
        int pivot = searchPivot(nums);

        // Step 2: If no pivot, perform regular binary search
        if (pivot == -1) {
            return binarySearch(nums, target, 0, nums.length - 1);
        }

        // Step 3: If the target is at the pivot, return the pivot index
        if (nums[pivot] == target) {
            return pivot;
        }

        // Step 4: Decide which half of the array to search
        if (target >= nums[0]) {
            return binarySearch(nums, target, 0, pivot - 1); // Search left side
        } else {
            return binarySearch(nums, target, pivot + 1, nums.length - 1); // Search right side
        }
    }

    // Binary search helper function
    static int binarySearch(int[] arr, int target, int start, int end) {
        while (start <= end) {
            int mid = start + (end - start) / 2;

            if (arr[mid] == target) {
                return mid; // Target found
            } else if (target < arr[mid]) {
                end = mid - 1; // Search left half
            } else {
                start = mid + 1; // Search right half
            }
        }
        return -1; // Target not found
    }

    // Function to find the pivot index in a rotated sorted array
    static int searchPivot(int[] arr) {
        int start = 0;
        int end = arr.length - 1;

        while (start <= end) {
            int mid = start + (end - start) / 2;

            // Check if mid is the pivot point
            if (mid < end && arr[mid] > arr[mid + 1]) {
                return mid;
            }

            // Check if the pivot is before the mid
            if (mid > start && arr[mid] < arr[mid - 1]) {
                return mid - 1;
            }

            // Decide whether to move left or right
            if (arr[mid] <= arr[start]) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }

        return -1; // No pivot found (array is not rotated)
    }
}