二刷268

Author: diguage

Diff for: docs/0268-missing-number.adoc

@@ -29,9 +29,32 @@ Given an array containing _n_ distinct numbers taken from `0, 1, 2, ..., n`, fin
 Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
 
 
+== 思路分析
+
 [[src-0268]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0268_MissingNumber.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0268_MissingNumber_2.java[tag=answer]
+----
+--
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/missing-number/solutions/1085105/diu-shi-de-shu-zi-by-leetcode-solution-naow/[268. 丢失的数字 - 官方题解^]
+. https://leetcode.cn/problems/missing-number/solutions/1086545/gong-shui-san-xie-yi-ti-wu-jie-pai-xu-ji-te3s/[268. 丢失的数字 - 一题五解 :「排序」&「计数」&「原地哈希」&「数学」&「异或」^]
 

Diff for: logbook/202406.adoc

@@ -419,6 +419,11 @@
 |{doc_base_url}/0041-first-missing-positive.adoc[题解]
 |循环排序，思路有意思！
 
+|{counter:codes}
+|{leetcode_base_url}/missing-number/[268. Missing Number^]
+|{doc_base_url}/0268-missing-number.adoc[题解]
+|循环排序
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

Diff for: src/main/java/com/diguage/algo/leetcode/_0268_MissingNumber.java

@@ -39,6 +39,9 @@ public class _0268_MissingNumber {
      * Memory Usage: 38.9 MB, less than 100.00% of Java online submissions for Missing Number.
      *
      * Copy from: https://leetcode.com/problems/missing-number/solution/[Missing Number solution - LeetCode]
+     *
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2020-01-05 21:30
      */
     public int missingNumber(int[] nums) {
         int actualSum = 0;

Diff for: src/main/java/com/diguage/algo/leetcode/_0268_MissingNumber_2.java

@@ -0,0 +1,38 @@
+package com.diguage.algo.leetcode;
+
+
+public class _0268_MissingNumber_2 {
+  // tag::answer[]
+
+  /**
+   * 循环排序的解法
+   *
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-08-29 19:07:11
+   */
+  public int missingNumber(int[] nums) {
+    int n = nums.length;
+    for (int i = 0; i < n; i++) {
+      if (nums[i] != i && nums[i] < n) {
+        swap(nums, nums[i], i--);
+      }
+    }
+    for (int i = 0; i < n; i++) {
+      if (nums[i] != i) {
+        return i;
+      }
+    }
+    return n;
+  }
+
+  private void swap(int[] nums, int i, int j) {
+    int temp = nums[i];
+    nums[i] = nums[j];
+    nums[j] = temp;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0268_MissingNumber_2().missingNumber(new int[]{1});
+  }
+}