|
| 1 | +/** |
| 2 | + * [137] Single Number II |
| 3 | + * |
| 4 | + * Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one. |
| 5 | + * |
| 6 | + * Note: |
| 7 | + * |
| 8 | + * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? |
| 9 | + * |
| 10 | + * Example 1: |
| 11 | + * |
| 12 | + * |
| 13 | + * Input: [2,2,3,2] |
| 14 | + * Output: 3 |
| 15 | + * |
| 16 | + * |
| 17 | + * Example 2: |
| 18 | + * |
| 19 | + * |
| 20 | + * Input: [0,1,0,1,0,1,99] |
| 21 | + * Output: 99 |
| 22 | + * |
| 23 | + */ |
| 24 | +pub struct Solution {} |
| 25 | + |
| 26 | +// submission codes start here |
| 27 | + |
| 28 | +/* |
| 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 |
| 30 | +
|
| 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 |
| 32 | +
|
| 33 | + 这题的真值表(3种状态使用2位): |
| 34 | +
|
| 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 |
| 39 | +
|
| 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 |
| 41 | +
|
| 42 | + 令 a' = 1, 则: |
| 43 | +
|
| 44 | + a b c a' |
| 45 | + 0 1 1 1 ~a & b & c |
| 46 | + 1 0 0 1 a & ~b & ~c |
| 47 | +
|
| 48 | + a' = (~a & b & c) | (a & ~b & ~c) |
| 49 | +
|
| 50 | + 同理: |
| 51 | +
|
| 52 | + b' = (~a & b & ~c) | (~a & ~b & c) |
| 53 | +
|
| 54 | + 这个每轮计算的位次数达到 17 次, 可以再优化一下: |
| 55 | +
|
| 56 | + 对 b' 化简: b' = ~a & (b & ~c | ~b & c) = ~a & b ^ c |
| 57 | +
|
| 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: |
| 59 | +
|
| 60 | + a (b) b' c a' b' |
| 61 | + 1 (0) 0 0 1 0 |
| 62 | + 0 (1) 0 1 1 0 |
| 63 | +
|
| 64 | + 重写一下就是 a' = (a & ~b' & ~c) | (~a & ~b' & c) = ~b' & (a & ~c | ~a & c) = ~b' & a ^ c |
| 65 | +
|
| 66 | + 这个就和最开始第二链接里给出的超简洁解法一致了 |
| 67 | +
|
| 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 | + */ |
| 70 | +impl Solution { |
| 71 | + pub fn single_number(nums: Vec<i32>) -> i32 { |
| 72 | + let (mut a, mut b) = (0, 0); |
| 73 | + for &num in nums.iter() { |
| 74 | + b = !a & (b ^ num); |
| 75 | + a = !b & (a ^ num); |
| 76 | + } |
| 77 | + return a | b |
| 78 | + } |
| 79 | +} |
| 80 | + |
| 81 | +// submission codes end |
| 82 | + |
| 83 | +#[cfg(test)] |
| 84 | +mod tests { |
| 85 | + use super::*; |
| 86 | + |
| 87 | + #[test] |
| 88 | + fn test_137() { |
| 89 | + assert_eq!(Solution::single_number(vec![0,0,0,1,1,1,5]), 5); |
| 90 | + } |
| 91 | +} |
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