solve: 136. Single Number

C solution with XOR operator

c/136.single-number/136.single-number.org

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+#+title: 136 Single Number
+
+Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
+
+You must implement a solution with a linear runtime complexity and use only constant extra space.
+
+
+
+Example 1:
+
+Input: nums = [2,2,1]
+
+Output: 1
+
+Example 2:
+
+Input: nums = [4,1,2,1,2]
+
+Output: 4
+
+Example 3:
+
+Input: nums = [1]
+
+Output: 1
+
+
+
+Constraints:
+
+    1 <= nums.length <= 3 * 104
+    -3 * 104 <= nums[i] <= 3 * 104
+    Each element in the array appears twice except for one element which appears only once.
+
+* Solution
+
+#+name: solution
+#+begin_src C
+int xor(int a, int b) {
+   return a ^ b;
+}
+
+int singleNumber(int* nums, int numsSize) {
+    int number = 0;
+    for (int i=0; i < numsSize; i++) {
+        number = xor(number, nums[i]);
+    }
+    return number;
+}
+#+end_src
+
+#+RESULTS: solution
+
+#+begin_src C :noweb yes
+#include <stdio.h>
+
+<<solution>>
+
+int numbers[3] = {1, 1, 2};
+int single = singleNumber(numbers, 3);
+printf("%d\n", single);
+#+end_src
+
+#+RESULTS:
+: -2