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Commit 2369db6

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Solve #260
1 parent e56d1eb commit 2369db6

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3 files changed

+70
-53
lines changed

3 files changed

+70
-53
lines changed

‎src/n0136_single_number.rs

Lines changed: 14 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -2,32 +2,32 @@
22
* [136] Single Number
33
*
44
* Given a non-empty array of integers, every element appears twice except for one. Find that single one.
5-
*
5+
*
66
* Note:
7-
*
7+
*
88
* Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
9-
*
9+
*
1010
* Example 1:
11-
*
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*
11+
*
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*
1313
* Input: [2,2,1]
1414
* Output: 1
15-
*
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*
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*
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*
1717
* Example 2:
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*
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*
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*
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*
2020
* Input: [4,1,2,1,2]
2121
* Output: 4
22-
*
23-
*
22+
*
23+
*
2424
*/
2525
pub struct Solution {}
2626

2727
// submission codes start here
2828
impl Solution {
2929
pub fn single_number(nums: Vec<i32>) -> i32 {
30-
nums.iter().fold(0, |acc, &num| {acc ^ num})
30+
nums.iter().fold(0, |acc, &num| acc ^ num)
3131
}
3232
}
3333

@@ -39,7 +39,7 @@ mod tests {
3939

4040
#[test]
4141
fn test_136() {
42-
assert_eq!(Solution::single_number(vec![2,2,1]), 1);
43-
assert_eq!(Solution::single_number(vec![4,1,2,1,2]), 4);
42+
assert_eq!(Solution::single_number(vec![2,2,1]), 1);
43+
assert_eq!(Solution::single_number(vec![4,1,2,1,2]), 4);
4444
}
4545
}

‎src/n0137_single_number_ii.rs

Lines changed: 37 additions & 37 deletions
Original file line numberDiff line numberDiff line change
@@ -2,79 +2,79 @@
22
* [137] Single Number II
33
*
44
* Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
5-
*
5+
*
66
* Note:
7-
*
7+
*
88
* Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
9-
*
9+
*
1010
* Example 1:
11-
*
12-
*
11+
*
12+
*
1313
* Input: [2,2,3,2]
1414
* Output: 3
15-
*
16-
*
15+
*
16+
*
1717
* Example 2:
18-
*
19-
*
18+
*
19+
*
2020
* Input: [0,1,0,1,0,1,99]
2121
* Output: 99
22-
*
22+
*
2323
*/
2424
pub struct Solution {}
2525

2626
// submission codes start here
2727

2828
/*
29-
糅合了一下 https://leetcode.com/problems/single-number-ii/discuss/43296/An-General-Way-to-Handle-All-this-sort-of-questions. 和 https://leetcode.com/problems/single-number-ii/discuss/43294/Challenge-me-thx
29+
糅合了一下 https://leetcode.com/problems/single-number-ii/discuss/43296/An-General-Way-to-Handle-All-this-sort-of-questions. 和 https://leetcode.com/problems/single-number-ii/discuss/43294/Challenge-me-thx
3030
31-
第一个链接给出了通用解法: 对于一个数出现 M 次其它数都出现了 K 的场景, 我们可以用位运算记录 K 种状态(作为一个计数器)来解
31+
第一个链接给出了通用解法: 对于一个数出现 M 次其它数都出现了 K 的场景, 我们可以用位运算记录 K 种状态(作为一个计数器)来解
3232
33-
这题的真值表(3种状态使用2位):
33+
这题的真值表(3种状态使用2位):
3434
35-
a b c/c a'b'/a'b'
36-
0 0 1/0 0 1 /0 0
37-
0 1 1/0 1 0 /0 1
38-
1 0 1/0 0 0 /1 0
35+
a b c/c a'b'/a'b'
36+
0 0 1/0 0 1 /0 0
37+
0 1 1/0 1 0 /0 1
38+
1 0 1/0 0 0 /1 0
3939
40-
根据数电的知识, 要根据这个真值表写出逻辑表达式, 以输出端为 '1' 的结果为准, 将每行的输入变量写成 AND 形式, 其中为 0 的输入量需要取反, 再将这几个 AND 形式做 OR 即可
40+
根据数电的知识, 要根据这个真值表写出逻辑表达式, 以输出端为 '1' 的结果为准, 将每行的输入变量写成 AND 形式, 其中为 0 的输入量需要取反, 再将这几个 AND 形式做 OR 即可
4141
42-
令 a' = 1, 则:
42+
令 a' = 1, 则:
4343
44-
a b c a'
45-
0 1 1 1 ~a & b & c
46-
1 0 0 1 a & ~b & ~c
44+
a b c a'
45+
0 1 1 1 ~a & b & c
46+
1 0 0 1 a & ~b & ~c
4747
48-
a' = (~a & b & c) | (a & ~b & ~c)
48+
a' = (~a & b & c) | (a & ~b & ~c)
4949
50-
同理:
50+
同理:
5151
52-
b' = (~a & b & ~c) | (~a & ~b & c)
52+
b' = (~a & b & ~c) | (~a & ~b & c)
5353
54-
这个每轮计算的位次数达到 17 次, 可以再优化一下:
54+
这个每轮计算的位次数达到 17 次, 可以再优化一下:
5555
56-
对 b' 化简: b' = ~a & (b & ~c | ~b & c) = ~a & b ^ c
56+
对 b' 化简: b' = ~a & (b & ~c | ~b & c) = ~a & b ^ c
5757
58-
但这时 a 仍然比较复杂, 我们可以考虑能否用每轮算出的 b' 来简化 a 的计算, 则:
58+
但这时 a 仍然比较复杂, 我们可以考虑能否用每轮算出的 b' 来简化 a 的计算, 则:
5959
60-
a (b) b' c a' b'
61-
1 (0) 0 0 1 0
62-
0 (1) 0 1 1 0
60+
a (b) b' c a' b'
61+
1 (0) 0 0 1 0
62+
0 (1) 0 1 1 0
6363
64-
重写一下就是 a' = (a & ~b' & ~c) | (~a & ~b' & c) = ~b' & (a & ~c | ~a & c) = ~b' & a ^ c
64+
重写一下就是 a' = (a & ~b' & ~c) | (~a & ~b' & c) = ~b' & (a & ~c | ~a & c) = ~b' & a ^ c
6565
66-
这个就和最开始第二链接里给出的超简洁解法一致了
66+
这个就和最开始第二链接里给出的超简洁解法一致了
6767
68-
最后的话, a 或 b 为 1 都可以输出到 1 (目标数出现1次或出现2次), 输出 a | b 即可
69-
*/
68+
最后的话, a 或 b 为 1 都可以输出到 1 (目标数出现1次或出现2次), 输出 a | b 即可
69+
*/
7070
impl Solution {
7171
pub fn single_number(nums: Vec<i32>) -> i32 {
7272
let (mut a, mut b) = (0, 0);
7373
for &num in nums.iter() {
7474
b = !a & (b ^ num);
7575
a = !b & (a ^ num);
7676
}
77-
return a | b
77+
return a | b;
7878
}
7979
}
8080

@@ -86,6 +86,6 @@ mod tests {
8686

8787
#[test]
8888
fn test_137() {
89-
assert_eq!(Solution::single_number(vec![0,0,0,1,1,1,5]), 5);
89+
assert_eq!(Solution::single_number(vec![0,0,0,1,1,1,5]), 5);
9090
}
9191
}

‎src/n0260_single_number_iii.rs

Lines changed: 19 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -22,7 +22,21 @@ pub struct Solution {}
2222

2323
impl Solution {
2424
pub fn single_number(nums: Vec<i32>) -> Vec<i32> {
25-
vec![]
25+
let mut res = 0;
26+
for &num in nums.iter() {
27+
res = res ^ num;
28+
}
29+
let right_most_set_bit = res & !(res - 1);
30+
let mut bit_set = 0;
31+
let mut bit_unset = 0;
32+
for &num in nums.iter() {
33+
if num & right_most_set_bit == 0 {
34+
bit_unset = bit_unset ^ num;
35+
} else {
36+
bit_set = bit_set ^ num;
37+
}
38+
}
39+
return vec![bit_set, bit_unset];
2640
}
2741
}
2842

@@ -33,5 +47,8 @@ mod tests {
3347
use super::*;
3448

3549
#[test]
36-
fn test_260() {}
50+
fn test_260() {
51+
assert_eq!(Solution::single_number(vec![1,2,1,2,3,4]), vec![3,4]);
52+
assert_eq!(Solution::single_number(vec![1,2,1,3,2,5]), vec![3,5]);
53+
}
3754
}

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