LeetCode/src/medium/SearchinRotatedSortedArrayII.java at master · chenyiAlone/LeetCode · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
package medium;
/**
 *
 * ClassName: SearchinRotatedSortedArrayII
 * @author chenyiAlone
 * Create Time: 2019/04/09 18:53:31
 * Description: No.81
 * 思路：
 * 一、
 *     1. 判断 mid 落在了那个区间
 *     2. 区间的种类
 *          [2, 5, 6, 0, 0, 1, 2]
 *           l                 h
 *           |  m  |                 hi: target < A[mid] && target >= A[lo]
 *                    |   m    |     lo: target > A[mid] && target <= A[hi]
 *
 *
 *
 * 二、
 *
 *     1. 数组是在一个位置打乱的，根据 nums[mid] 和 nums[lo] 、nums[hi] 大小关系分为两部分
 *         1). nums[mid] < nums[hi] || nums[mid] < nums[lo] , (oldStart 和 lo 可能重合)
 *                 对应 { nums[lo], ... , nums[oldStart], ..., nums[mid], ..., nums[hi] }
 *                 lo 右移的唯一情况是 target < nums[mid] && target  <= nums[hi]
 *
 *         2). nums[mid] > nums[lo] || nums[mid] > nums[hi] , (oldStart 和 hi 可能重合)
 *                 对应 { nums[lo], ... , nums[mid], ... , nums[oldStart], ... , nums[hi] }
 *                 hi 左移的唯一情况是 nums[mid] < target && nums[lo] <= target
 *
 * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

    (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

    You are given a target value to search. If found in the array return true, otherwise return false.

    Example 1:

    Input: nums = [2,5,6,0,0,1,2], target = 0
    Output: true
    Example 2:

    Input: nums = [2,5,6,0,0,1,2], target = 3
    Output: false
    Follow up:

    This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
    Would this affect the run-time complexity? How and why?
 */
public class SearchinRotatedSortedArrayII {
    public boolean search(int[] nums, int target) {
        int lo = 0, hi = nums.length - 1;
         while (lo <= hi) {
             int mid = (lo + hi) >> 1;
             if (nums[mid] == target)
                 return true;
             if (nums[mid] < nums[hi] || nums[mid] < nums[lo]) {
                 if (target > nums[mid] && target <= nums[hi])
                     lo = mid + 1;
                 else
                     hi = mid - 1;

             } else if (nums[mid] > nums[lo] || nums[mid] > nums[hi]) {
                 if (target < nums[mid] && target >= nums[lo])
                     hi = mid - 1;
                 else
                     lo = mid + 1;
             } else {
                 hi--;
             }
         }
         return false;
     }

    public boolean search2(int[] nums, int target) {
        int lo = 0, hi = nums.length - 1;
        while (lo <= hi) {
            int mid = (lo + hi) >> 1;
            if (nums[mid] == target)
                return true;
            if (nums[mid] < nums[hi] || nums[mid] < nums[lo]) {
                if (target > nums[mid] && target <= nums[hi])
                    lo = mid + 1;
                else
                    hi = mid - 1;

            } else if (nums[mid] > nums[lo] || nums[mid] > nums[hi]) {
                if (target < nums[mid] && target >= nums[lo])
                    hi = mid - 1;
                else
                    lo = mid + 1;
            } else {
                hi--;
            }
        }
        return false;
    }
 }