二刷27

Author: diguage

docs/0027-remove-element.adoc

@@ -1,63 +1,89 @@
 [#0027-remove-element]
-= 27. Remove Element
+= 27. 移除元素
 
-{leetcode}/problems/remove-element/[LeetCode - Remove Element^]
+https://leetcode.cn/problems/remove-element/[LeetCode - 27. 移除元素 ^]
 
-Given an array _nums_ and a value _val_, remove all instances of that value https://en.wikipedia.org/wiki/In-place_algorithm[*in-place*^] and return the new length.
+给你一个数组 `nums` 和一个值 `val`，你需要 *https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95[原地]* 移除所有数值等于 `val` 的元素。元素的顺序可能发生改变。然后返回 `nums` 中与 `val` 不同的元素的数量。
 
-Do not allocate extra space for another array, you must do this by *modifying the input array https://en.wikipedia.org/wiki/In-place_algorithm[in-place^]* with O(1) extra memory.
+假设 `nums` 中不等于 `val` 的元素数量为 `k`，要通过此题，您需要执行以下操作：
 
-The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+* 更改 `nums` 数组，使 `nums` 的前 `k` 个元素包含不等于 `val` 的元素。`nums` 的其余元素和 `nums` 的大小并不重要。* 返回 `k`。
 
-*Example 1:*
+*用户评测：*
 
-[subs="verbatim,quotes,macros"]
-----
-Given _nums_ = *[3,2,2,3]*, _val_ = *3*,
-
-Your function should return length = *2*, with the first two elements of _nums_ being *2*.
+评测机将使用以下代码测试您的解决方案：
 
-It doesn't matter what you leave beyond the returned length.
+....
+int[] nums = [...]; // 输入数组
+int val = ...; // 要移除的值
+int[] expectedNums = [...]; // 长度正确的预期答案。
+                            // 它以不等于 val 的值排序。
 
-----
+int k = removeElement(nums, val); // 调用你的实现
 
-*Example 2:*
+assert k == expectedNums.length;
+sort(nums, 0, k); // 排序 nums 的前 k 个元素
+for (int i = 0; i < actualLength; i++) {
+    assert nums[i] == expectedNums[i];
+}
+....
 
-[subs="verbatim,quotes,macros"]
-----
-Given _nums_ = *[0,1,2,2,3,0,4,2]*, _val_ = *2*,
+如果所有的断言都通过，你的解决方案将会 *通过*。
 
-Your function should return length = *`5`*, with the first five elements of _`nums`_ containing *`0`*, *`1`*, *`3`*, *`0`*, and *4*.
 
-Note that the order of those five elements can be arbitrary.
+*示例 1：*
 
-It doesn't matter what values are set beyond the returned length.
-----
+....
+输入：nums = [3,2,2,3], val = 3
+输出：2, nums = [2,2,_,_]
+解释：你的函数函数应该返回 k = 2, 并且 nums 中的前两个元素均为 2。
+你在返回的 k 个元素之外留下了什么并不重要（因此它们并不计入评测）。
+....
 
-*Clarification:*
+*示例 2：*
 
-Confused why the returned value is an integer but your answer is an array?
+....
+输入：nums = [0,1,2,2,3,0,4,2], val = 2
+输出：5, nums = [0,1,4,0,3,_,_,_]
+解释：你的函数应该返回 k = 5，并且 nums 中的前五个元素为 0,0,1,3,4。
+注意这五个元素可以任意顺序返回。
+你在返回的 k 个元素之外留下了什么并不重要（因此它们并不计入评测）。
+....
 
-Note that the input array is passed in by *reference*, which means modification to the input array will be known to the caller as well.
+*提示：*
 
-Internally you can think of this:
+* `+0 <= nums.length <= 100+`
+* `+0 <= nums[i] <= 50+`
+* `+0 <= val <= 100+`
 
-[subs="verbatim,quotes,macros"]
-----
-// *nums* is passed in by reference. (i.e., without making a copy)
-int len = removeElement(nums, val);
 
-// any modification to *nums* in your function would be known by the caller.
-// using the length returned by your function, it prints the first *len* elements.
-for (int i = 0; i < len; i++) {
-    print(nums[i]);
-}
-----
+== 思路分析
 
+快慢指针。慢指针记录满足条件的长度，快指针指向需要处理的元素。如果 `nums[fast] == val`，则只移动 `fast`；否则，`nums[slow++] == nums[fast++]`。
 
 [[src-0027]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0027_RemoveElement.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0027_RemoveElement_2.java[tag=answer]
+----
+--
+====
+
+
+== 参考资料
+
 

logbook/202503.adoc

@@ -363,6 +363,11 @@ endif::[]
 |{doc_base_url}/0011-container-with-most-water.adoc[题解]
 |✅ 双指针
 
+|{counter:codes}
+|{leetcode_base_url}/remove-element/[27. 移除元素^]
+|{doc_base_url}/0027-remove-element.adoc[题解]
+|✅ 快慢指针。慢指针记录满足条件的长度，快指针指向需要处理的元素。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0027_RemoveElement.java

@@ -62,6 +62,11 @@
  */
 public class _0027_RemoveElement {
   // tag::answer[]
+
+    /**
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2018-07-19 18:55
+     */
     public static int removeElement(int[] nums, int val) {
         if (nums == null || nums.length == 0) {
             return 0;

src/main/java/com/diguage/algo/leetcode/_0027_RemoveElement_2.java

@@ -0,0 +1,21 @@
+package com.diguage.algo.leetcode;
+
+public class _0027_RemoveElement_2 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-20 18:25:23
+   */
+  public static int removeElement(int[] nums, int val) {
+    int slow = 0, fast = 0;
+    while (fast < nums.length) {
+      if (nums[fast] == val) {
+        fast++;
+      }else {
+        nums[slow++] = nums[fast++];
+      }
+    }
+    return slow;
+  }
+  // end::answer[]
+}