leetcode 81. 搜索旋转排序数组 II

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。

( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。

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。

示例 1:

输入: nums = [2,5,6,0,0,1,2], target = 0
输出: true
示例 2:

输入: nums = [2,5,6,0,0,1,2], target = 3
输出: false

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 1 public class _81 {
 2     public boolean search(int[] nums, int target) {
 3         if (nums.length == 2) return nums[0] == target || nums[1] == target;
 4         if (nums.length == 1) return nums[0] == target;
 5         if (nums.length == 0) return false;
 6         int low = 0, high = nums.length-1, mid = (low+high)/2;
 7         int nsl = 0, nsr = high;
 8 
 9         if (nums[low] == nums[mid] && nums[mid] == nums[high]) {
10             for (int i = 0; i < nums.length; i++)
11                 if (nums[i] == target)
12                     return true;
13             return false;
14         }
15 
16         // 找出有序的那部分
17         int[] t = null;
18         if (nums[mid] <= nums[high]){ // 左边旋转过，右边有序
19             t = Arrays.copyOfRange(nums, mid, high+1);
20             nsl = low;
21             nsr = mid+1;
22         } else {
23             t = Arrays.copyOfRange(nums, low, mid+1);
24             nsl = mid;
25             nsr = high+1;
26         }
27         int res = Arrays.binarySearch(t, target);
28         if (res >= 0) return true;
29 
30         // 判断无序的左边是否存在target
31         return search(Arrays.copyOfRange(nums, nsl, nsr), target);
32     }
33 
34     public static void main(String[] args) {
35         int[] nums = {1,1,1,1,1,1,1,2,1,1};
36         boolean res = new _81().search(nums, 2);
37         System.out.println("res = " + res);
38     }
39 }

稍微改进下。

 1 public class _81 {
 2     public boolean binarySearch(int[] nums, int low, int high, int target){ // 包括high位置
 3         int mid;
 4 
 5         while(low <= high){
 6             mid = (low+high)/2;
 7             if (nums[mid] < target){
 8                 low = mid+1;
 9             } else if (nums[mid] > target){
10                 high = mid - 1;
11             } else {
12                 return true;
13             }
14         }
15         return false;
16     }
17     public boolean result(int[] nums, int target, int low, int high){
18         if (low+1==high) return nums[low] == target || nums[high] == target;
19         if (low == high) return nums[low] == target;
20         int mid = (low+high)/2;
21 
22         if (nums[low] == nums[mid] && nums[mid] == nums[high]) {
23             for (int i = low; i <= high; i++)
24                 if (nums[i] == target)
25                     return true;
26             return false;
27         }
28 
29         if (nums[mid] <= nums[high]){ // 右边有序，且taregt存在于右边
30             if (target <= nums[high] && target >= nums[mid])
31                 return binarySearch(nums, mid, high, target);
32             // 查找无序的左边
33             return result(nums, target, low, mid);
34         }
35         if (nums[low] <= nums[mid]){ // 左边有序，且target存在于左边
36             if (target <= nums[mid] && target >= nums[low])
37                 return binarySearch(nums, low, mid, target);
38             // 查找无序的右边
39             return result(nums, target, mid, high);
40         }
41         return false;
42     }
43     public boolean search(int[] nums, int target) {
44         if (nums.length == 2) return nums[0] == target || nums[1] == target;
45         if (nums.length == 1) return nums[0] == target;
46         if (nums.length == 0) return false;
47         return result(nums, target, 0, nums.length-1);
48     }
49 
50     public static void main(String[] args) {
51         int[] nums = {2,5,6,0,0,1,2};
52         boolean res = new _81().search(nums, 8);
53         System.out.println("res = " + res);
54     }
55 }