代码挑战：找到单一数字|Rust
【摘要】 Given a non-empty array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without usi...
Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4方案1：
impl Solution {
pub fn single_number(nums: Vec<i32>) -> i32 {
let mut result = nums[0];
for (pos, e) in nums.iter().enumerate() {
if pos != 0 {
result ^= e;
}
}
return result;
}
}
方案2：
impl Solution {
pub fn single_number(nums: Vec<i32>) -> i32 {
let mut hs = std::collections::HashSet::new();
for (_pos, e) in nums.iter().enumerate() {
if hs.contains(e) {
hs.remove(&e);
}
else {
hs.insert(e);
}
}
for x in hs.iter() {
return **x;
}
return 0;
}
}
测试案例：
#[test]
fn test1() {
let vec = vec![2,2,1];
let result = Solution::single_number(vec);
assert_eq!(result, 1);
}
#[test]
fn test2() {
let vec = vec![4, 1, 2, 1, 2];
let result = Solution::single_number(vec);
assert_eq!(result, 4);
}
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