Bonus puzzle for April challenge

136/main.rs

@@ -0,0 +1,22 @@
+use std::collections::HashMap;
+
+// I'm fairly sure there's a better more mathsy way of doing this but I can't work it out yet.
+pub fn single_number(nums: Vec<i32>) -> i32 {
+    let mut hm = HashMap::<i32, i32>::new();
+    for n in nums {
+        if hm.contains_key(&n) {
+            hm.remove(&n);
+        } else {
+            hm.insert(n, 0);
+        }
+    } 
+    match hm.keys().next() {
+        Some(i) => *i,
+        _ => 0
+    }
+}
+
+pub fn main() {
+    println!("Expected: 1");
+    println!("Got: {}", single_number(vec![1, 2, 3, 4, 3, 4, 2]));
+}

136/problem.txt

@@ -0,0 +1,15 @@
+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

136/run.sh

@@ -0,0 +1,5 @@
+#!/bin/bash
+
+rustc main.rs
+./main
+rm main