auto commit

notes/56. 数组中只出现一次的数字.md

@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/e02fdb54d7524710a7d664d082bb7811?tpId=13&tqId=11193&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/389fc1c3d3be4479a154f63f495abff8?tpId=13&tqId=11193&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -19,16 +19,27 @@
 下面的解法中，num1 和 num2 数组的第一个元素是用来保持返回值的... 实际开发中不推荐这种返回值的方式。
 
 ```java
-public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
+public int[] FindNumsAppearOnce (int[] nums) {
+    int[] res = new int[2];
     int diff = 0;
     for (int num : nums)
         diff ^= num;
     diff &= -diff;
     for (int num : nums) {
         if ((num & diff) == 0)
-            num1[0] ^= num;
+            res[0] ^= num;
         else
-            num2[0] ^= num;
+            res[1] ^= num;
     }
+    if (res[0] > res[1]) {
+        swap(res);
+    }
+    return res;
+}
+
+private void swap(int[] nums) {
+    int t = nums[0];
+    nums[0] = nums[1];
+    nums[1] = t;
 }
 ```