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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=81 lang=java |
| 3 | + * |
| 4 | + * [81] 搜索旋转排序数组 II |
| 5 | + * |
| 6 | + * https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Medium (35.56%) |
| 10 | + * Likes: 187 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 32.3K |
| 13 | + * Total Submissions: 90.3K |
| 14 | + * Testcase Example: '[2,5,6,0,0,1,2]\n0' |
| 15 | + * |
| 16 | + * 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 |
| 17 | + * |
| 18 | + * ( 例如,数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。 |
| 19 | + * |
| 20 | + * 编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true,否则返回 false。 |
| 21 | + * |
| 22 | + * 示例 1: |
| 23 | + * |
| 24 | + * 输入: nums = [2,5,6,0,0,1,2], target = 0 |
| 25 | + * 输出: true |
| 26 | + * |
| 27 | + * |
| 28 | + * 示例 2: |
| 29 | + * |
| 30 | + * 输入: nums = [2,5,6,0,0,1,2], target = 3 |
| 31 | + * 输出: false |
| 32 | + * |
| 33 | + * 进阶: |
| 34 | + * |
| 35 | + * |
| 36 | + * 这是 搜索旋转排序数组 的延伸题目,本题中的 nums 可能包含重复元素。 |
| 37 | + * 这会影响到程序的时间复杂度吗?会有怎样的影响,为什么? |
| 38 | + * |
| 39 | + * |
| 40 | + */ |
| 41 | + |
| 42 | +// @lc code=start |
| 43 | +class Solution { |
| 44 | + public boolean search(int[] nums, int target) { |
| 45 | + int left = 0; |
| 46 | + int right = nums.length - 1; |
| 47 | + |
| 48 | + while (left <= right) { |
| 49 | + int mid = left + (right - left) / 2; |
| 50 | + if (nums[mid] == target) { |
| 51 | + return true; |
| 52 | + } |
| 53 | + if (nums[left] == nums[mid]) { |
| 54 | + left++; |
| 55 | + } else if (nums[left] < nums[mid]) { |
| 56 | + if (nums[left] <= target && target < nums[mid]) { |
| 57 | + right = mid - 1; |
| 58 | + } else { |
| 59 | + left = mid + 1; |
| 60 | + } |
| 61 | + } else if (nums[left] > nums[mid]) { |
| 62 | + if (nums[mid] < target && target <= nums[right]) { |
| 63 | + left = mid + 1; |
| 64 | + } else { |
| 65 | + right = mid - 1; |
| 66 | + } |
| 67 | + } |
| 68 | + } |
| 69 | + |
| 70 | + return false; |
| 71 | + } |
| 72 | +} |
| 73 | +// @lc code=end |
| 74 | + |
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