Leetcode Search in Rotated Sorted Array

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本文介绍了一种在旋转数组中寻找目标值的算法实现。通过两种方法解决该问题：一种是简单遍历查找，时间复杂度为O(n)；另一种是采用二分查找法，将时间复杂度降低到O(log n)。文章提供了完整的Java代码示例。

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Suppose a sorted array 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.

Difficulty: Hard

Solution 1: O(n), the simplest way.

public class Solution {
    public int search(int[] nums, int target) {
        int i = 0;
        int len = nums.length;
        while(i < len){
            if(target == nums[i])
                return i;
            i++;
        }
        return -1;
    }
}

Solution 2: O(logn), Binary Search.

public class Solution {
    public int search(int[] nums, int target) {
        return binSearch(nums, 0, nums.length-1, target);
    }
    public int binSearch(int[] nums, int start, int end, int target){
        if(start > end)
            return -1;
        int mid = start + (end-start)/2;
        
        if(nums[mid] == target)
            return mid;
        
        if(nums[mid] >= nums[start]){
            if(target >= nums[start] && target < nums[mid]){
                return binSearch(nums, start, mid - 1, target);
            }
            else{
                return binSearch(nums, mid + 1, end, target);
            }
        }
        else{
            if(target <= nums[end] && target > nums[mid]){
                return binSearch(nums, mid + 1, end, target);
            }
            else{
                return binSearch(nums, start, mid - 1, target);
            }
        }
    }
}

Solution 1: O(n), the simplest way.