算法：搜索数字，从有逆序的顺序数组33. Search in Rotated Sorted Array

Posted 2021-09-01 架构师易筋

33. Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

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

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is guaranteed to be rotated at some pivot.
-104 <= target <= 104

解法

想象一下二分法，判断有序的部分，

1. 如果target在于 min < target < max 中，则直接在这份范围里找，
2. 否则肯定在另一部分。

以此类推

class Solution {
    public int search(int[] nums, int target) {
        return helper(nums, target, 0, nums.length -1);
    }
    
    private int helper(int[] nums, int target, int l, int r) {
        if (l > r) {
            return -1;
        }
        
        int mid = l + ((r - l) >> 1);
        if (nums[mid] == target) {
            return mid;
        }
        
        if (nums[l] <= nums[mid]) {
            if (nums[mid] > target && nums[l] <= target) {
                return helper(nums, target, l, mid - 1);
            } else {
                return helper(nums, target, mid + 1, r);
            }
        } else {
            if (nums[r] >= target && nums[mid] < target) {
                return helper(nums, target, mid + 1, r);
            } else {
                return helper(nums, target, l, mid - 1);
            }
        }
    }
}