[leetcode]33. Search in Rotated Sorted Array(Java实现)

测试地址：https://leetcode.com/problems/search-in-rotated-sorted-array/#/description

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.

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package go.jacob.day620;/** * [leetcode]33.Search in Rotated Sorted Array *  * @author Administrator * */public class Demo2 {public int search(int[] nums, int target) {if (nums == null || nums.length < 1)return -1;int result = searchTarget(nums, target, 0, nums.length - 1);return result;}/* * 循环 */private int searchTarget(int[] nums, int target, int left, int right) {while (left < right) {int mid = left + (right - left) / 2;if (nums[mid] == target)return mid;if (nums[mid] >= nums[left]) {// 考虑较简单的左半边情况，else都在右半边if (target >= nums[left] && target < nums[mid])right = mid - 1;elseleft = mid + 1;} else {if (target > nums[mid] && target <= nums[right])left = mid + 1;elseright = mid - 1;}}//当right=left+1时，mid=left，if (nums[mid] == target)return mid;不满足，只有可能是-1，或者left+1return nums[left] == target ? left : -1;}/* * 递归 */private int searchTarget_1(int[] nums, int target, int left, int right) {if (right < left)return -1;int mid = left + (right - left) / 2;if (nums[mid] == target)return mid;// 这里nums[mid]必须是大等于nums[left]if (nums[mid] >= nums[left]) {if (target >= nums[left] && target < nums[mid])return searchTarget(nums, target, left, mid - 1);elsereturn searchTarget(nums, target, mid + 1, right);} else {if (target > nums[mid] && target <= nums[right])return searchTarget(nums, target, mid + 1, right);elsereturn searchTarget(nums, target, left, mid - 1);}}}