Update 剑指 Offer 题解 - 3~9.md

notes/剑指 Offer 题解 - 3~9.md

@@ -58,7 +58,39 @@ private void swap(int[] nums, int i, int j) {
     nums[j] = t;
 }
 ```
+```java
+public class Solution {
+    // Parameters:
+    //    numbers:     an array of integers
+    //    length:      the length of array numbers
+    //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
+    //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
+    //    这里要特别注意~返回任意重复的一个，赋值duplication[0]
+    // Return value:       true if the input is valid, and there are some duplications in the array number
+    //                     otherwise false
+   public boolean duplicate(int[] numbers, int length, int[] duplication) {
+    if (numbers == null || length <= 0)
+        return false;
+    for (int i = 0; i < length; i++) {
+        while (numbers[i] != i) {
+            if (numbers[i] == numbers[numbers[i]]) {
+                duplication[0] = numbers[i];
+                return true;
+            }
+                int temp=numbers[i];
+                numbers[i]=numbers[temp];
+                numbers[temp]=temp;
+            //swap(numbers, i, numbers[i]);
+        }
+    }
+    return false;
+}
  
+private void swap(int[] nums, int i, int j) {
+    int t = nums[i]; nums[i] = nums[j]; nums[j] = t;
+}
+}
+```
 # 4. 二维数组中的查找
 
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