forked from haoel/leetcode
-
Notifications
You must be signed in to change notification settings - Fork 39
/
MissingNumber.cpp
54 lines (49 loc) · 1.78 KB
/
MissingNumber.cpp
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
// Source : https://leetcode.com/problems/missing-number/
// Author : Hao Chen
// Date : 2015-10-22
/***************************************************************************************
*
* Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the
* one that is missing from the array.
*
* For example,
* Given nums = [0, 1, 3] return 2.
*
* Note:
* Your algorithm should run in linear runtime complexity. Could you implement it using
* only constant extra space complexity?
*
* Credits:Special thanks to @jianchao.li.fighter for adding this problem and creating
* all test cases.
*
***************************************************************************************/
class Solution {
public:
// This problem can be converted to the classic problem --
// `There is an array, all of numbers except one appears twice, and that one only appears once`
// It means, we can combin two arrays together, one is [1..n], another one is `nums`.
// Then, you know, we can use the XOR solve this problem.
int missingNumber01(vector<int>& nums) {
int result = 0;
for(int i=0; i<nums.size(); i++){
result ^= nums[i];
}
for(int i=1; i<=nums.size(); i++){
result ^=(i);
}
return result;
}
// We can simplify the previous solution as below
int missingNumber02(vector<int>& nums) {
int result = 0;
for(int i=0; i<nums.size(); i++){
result = result ^ (i+1) ^ nums[i];
}
return result;
}
int missingNumber(vector<int>& nums) {
//By Leetcode running result, they all are same performance
return missingNumber02(nums); //36ms
return missingNumber01(nums); //36ms
}
};