Search in Rotated Sorted Array——LeetCode

 

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

题目大意：有序数组旋转，寻找其中是否有某个特定的值，时间复杂度O(log n)。

思路：旋转后的数组，旋转后的后一部分肯定比前一部分所有元素都小；所以数组从中间分开，至少有一半是有序的。

那么可以得到，如果中间的元素（也就是nums[mid]）比nums[0]大，那么前半部分有序，如果中间的元素比nums[len-1]小，那么后半部分有序，这两种情况只会出现一种。这样就可以先看这个数在不在有序的这一半数组中，在的话没什么好说的，二分查找；不在的话，继续找另一半RotatedArray。思路就是这么个思路，然而想把题目A掉还要考虑很多细节，尤其是边界值，> or >= ，< or <=等等，极易出错。

public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int low = 0, high = nums.length - 1;
        while (low <= high) {
            int mid = (low + high) >> 1;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[low] <= nums[mid]) {
                if (target >= nums[low] && target <= nums[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            } else {
                if (target >= nums[mid] && target <= nums[high]) {
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
        }
        return -1;
    }