-
Notifications
You must be signed in to change notification settings - Fork 1
/
Question070.java
77 lines (72 loc) · 2.29 KB
/
Question070.java
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
package com.hncboy.swordreferstoofferspecial;
import com.hncboy.SingleElementInASortedArray;
/**
* @author hncboy
* @date 2022/1/1 16:19
* 剑指 Offer II 070.排序数组中只出现一次的数字
*
* 给定一个只包含整数的有序数组 nums ,每个元素都会出现两次,唯有一个数只会出现一次,请找出这个唯一的数字。
*
* 示例 1:
* 输入: nums = [1,1,2,3,3,4,4,8,8]
* 输出: 2
*
* 示例 2:
* 输入: nums = [3,3,7,7,10,11,11]
* 输出: 10
*
* 提示:
* 1 <= nums.length <= 105
* 0 <= nums[i] <= 105
*
* 进阶: 采用的方案可以在 O(log n) 时间复杂度和 O(1) 空间复杂度中运行吗?
* 注意:本题与主站 540 题 {@link SingleElementInASortedArray} 相同:https://leetcode-cn.com/problems/single-element-in-a-sorted-array/
*
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/skFtm2
* 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。
*/
public class Question070 {
public int singleNonDuplicate2(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
// 确保 mid 是偶数
if (mid % 2 == 1) {
mid--;
}
if (nums[mid] == nums[mid + 1]) {
left = mid + 2;
} else {
right = mid;
}
}
return nums[left];
}
public int singleNonDuplicate1(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
// 判断出现单次的数字在哪一边
boolean halvesAreEven = (right - mid) % 2 == 0;
if (nums[mid + 1] == nums[mid]) {
if (halvesAreEven) {
left = mid + 2;
} else {
right = mid - 1;
}
} else if (nums[mid - 1] == nums[mid]) {
if (halvesAreEven) {
right = mid - 2;
} else {
left = mid + 1;
}
} else {
return nums[mid];
}
}
return nums[left];
}
}