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‎docs/0080-remove-duplicates-from-sorted-array-ii.adoc

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[#0080-remove-duplicates-from-sorted-array-ii]
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= 80. Remove Duplicates from Sorted Array II
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= 80. 删除有序数组中的重复项 II
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{leetcode}/problems/remove-duplicates-from-sorted-array-ii/[LeetCode - Remove Duplicates from Sorted Array II^]
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https://leetcode.cn/problems/remove-duplicates-from-sorted-array-ii/[LeetCode - 80. 删除有序数组中的重复项 II ^]
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Given a sorted array `nums`, remove the duplicates in-place such that duplicates appeared at most *twice* and return the new length.
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给你一个有序数组 `nums` ，请你 *http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95[原地]* 删除重复出现的元素，使得出现次数超过两次的元素**只出现两次**，返回删除后数组的新长度。
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Do not allocate extra space for another array, you must do this by *modifying the input array* in-place with O(1) extra memory.
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不要使用额外的数组空间，你必须在 *https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95[原地] 修改输入数组* 并在使用 O(1) 额外空间的条件下完成。
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.Example 1:
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----
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Given nums = [1,1,1,2,2,3],
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Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
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It doesn't matter what you leave beyond the returned length.
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----
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*说明：*
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.Example 2:
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----
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Given nums = [0,0,1,1,1,1,2,3,3],
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为什么返回数值是整数，但输出的答案是数组呢？
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Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.
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请注意，输入数组是以**「引用」**方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。
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It doesn't matter what values are set beyond the returned length.
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----
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你可以想象内部操作如下:
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*Clarification:*
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Confused why the returned value is an integer but your answer is an array?
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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.
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Internally you can think of this:
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----
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// nums is passed in by reference. (i.e., without making a copy)
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....
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// nums 是以“引用”方式传递的。也就是说，不对实参做任何拷贝
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int len = removeDuplicates(nums);
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// any modification to nums in your function would be known by the caller.
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// using the length returned by your function, it prints the first len elements.
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// 在函数里修改输入数组对于调用者是可见的。
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// 根据你的函数返回的长度, 它会打印出数组中 该长度范围内 的所有元素。
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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----
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....
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== 解题分析
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解题思路很简单：把后面的值覆盖前面多余的值。关键是如何用简单的代码来实现这个思路。
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*示例 1：*
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== 参考资料
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....
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输入：nums = [1,1,1,2,2,3]
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输出：5, nums = [1,1,2,2,3]
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解释：函数应返回新长度 length = 5, 并且原数组的前五个元素被修改为 1, 1, 2, 2, 3。 不需要考虑数组中超出新长度后面的元素。
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....
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. https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array-ii/solution/shan-chu-pai-xu-shu-zu-zhong-de-zhong-fu-xiang-i-7/[删除排序数组中的重复项 II - 删除排序数组中的重复项 II - 力扣（LeetCode）^]
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*示例 2：*
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Given a sorted array _nums_, remove the duplicates https://en.wikipedia.org/wiki/In-place_algorithm[*in-place*^] such that duplicates appeared at most _twice_ and return the new length.
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....
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输入：nums = [0,0,1,1,1,1,2,3,3]
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输出：7, nums = [0,0,1,1,2,3,3]
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解释：函数应返回新长度 length = 7, 并且原数组的前七个元素被修改为 0, 0, 1, 1, 2, 3, 3。不需要考虑数组中超出新长度后面的元素。
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....
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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.
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*Example 1:*
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*提示：*
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[subs="verbatim,quotes,macros"]
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----
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Given _nums_ = *[1,1,1,2,2,3]*,
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* `1 \<= nums.length \<= 3 * 10^4^`
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* `-10^4^ \<= nums[i] \<= 10^4^`
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* `nums` 已按升序排列
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Your function should return length = *`5`*, with the first five elements of _`nums`_ being *`1, 1, 2, 2`* and *3* respectively.
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It doesn't matter what you leave beyond the returned length.
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----
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== 思路分析
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*Example 2:*
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解题思路很简单：把后面的值覆盖前面多余的值。关键是如何用简单的代码来实现这个思路。
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[subs="verbatim,quotes,macros"]
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[[src-0080]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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Given _nums_ = *[0,0,1,1,1,1,2,3,3]*,
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Your function should return length = *`7`*, with the first seven elements of _`nums`_ being modified to *`0`*, *0*, *1*, *1*, *2*, *3* and *3* respectively.
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It doesn't matter what values are set beyond the returned length.
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include::{sourcedir}/_0080_RemoveDuplicatesFromSortedArrayII.java[tag=answer]
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----
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--
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*Clarification:*
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Confused why the returned value is an integer but your answer is an array?
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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.
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Internally you can think of this:
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[subs="verbatim,quotes,macros"]
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二刷::
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--
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[{java_src_attr}]
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----
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// *nums* is passed in by reference. (i.e., without making a copy)
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int len = removeDuplicates(nums);
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// any modification to *nums* in your function would be known by the caller.
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// using the length returned by your function, it prints the first *len* elements.
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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include::{sourcedir}/_0080_RemoveDuplicatesFromSortedArrayIi_2.java[tag=answer]
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----
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--
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====
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[[src-0080]]
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[{java_src_attr}]
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----
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include::{sourcedir}/_0080_RemoveDuplicatesFromSortedArrayII.java[tag=answer]
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----
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== 参考资料
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. https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array-ii/solution/shan-chu-pai-xu-shu-zu-zhong-de-zhong-fu-xiang-i-7/[删除排序数组中的重复项 II - 删除排序数组中的重复项 II - 力扣（LeetCode）^]

‎logbook/202503.adoc

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|{doc_base_url}/0045-jump-game-ii.adoc[题解]
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|✅ 贪心算法。也可以动态规划。
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|{counter:codes2503}
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|{leetcode_base_url}/remove-duplicates-from-sorted-array-ii/[80. 删除有序数组中的重复项 II^]
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|{doc_base_url}/0080-remove-duplicates-from-sorted-array-ii.adoc[题解]
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|✅ 双指针，同时需要记录每个元素的次数，相同元素只复制两个。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0080_RemoveDuplicatesFromSortedArrayII.java

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* Memory Usage: 41.7 MB, less than 5.26% of Java online submissions for Remove Duplicates from Sorted Array II.
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*
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* Copy from: https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array-ii/solution/shan-chu-pai-xu-shu-zu-zhong-de-zhong-fu-xiang-i-7/[删除排序数组中的重复项 II - 删除排序数组中的重复项 II - 力扣（LeetCode）]
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*
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* @author D瓜哥 · https://www.diguage.com
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* @since 2020-02-04 20:57
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*/
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public int removeDuplicates(int[] nums) {
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int j = 1, count = 1;

‎src/main/java/com/diguage/algo/leetcode/_0080_RemoveDuplicatesFromSortedArrayIi_2.java

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package com.diguage.algo.leetcode;
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public class _0080_RemoveDuplicatesFromSortedArrayIi_2 {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-18 15:08:39
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*/
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public int removeDuplicates(int[] nums) {
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int slow = 0, fast = 0, cnt = 1;
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while (fast < nums.length) {
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if (nums[fast] != nums[slow]) {
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cnt = 1;
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nums[++slow] = nums[fast];
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} else {
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if (cnt < 2 && slow != fast) {
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nums[++slow] = nums[fast];
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cnt++;
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}
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}
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fast++;
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}
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return slow + 1;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0080_RemoveDuplicatesFromSortedArrayIi_2()
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.removeDuplicates(new int[]{0, 0, 1, 1, 1, 1, 2, 3, 3});
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}
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}

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