题目地址
https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/description/
题目描述
给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素只出现一次，返回移除后数组的新长度。
不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。
示例 1:
给定数组 nums = [1,1,2],
函数应该返回新的长度 2, 并且原数组 nums 的前两个元素被修改为 1, 2。
你不需要考虑数组中超出新长度后面的元素。
示例 2:
给定 nums = [0,0,1,1,1,2,2,3,3,4],
函数应该返回新的长度 5, 并且原数组 nums 的前五个元素被修改为 0, 1, 2, 3, 4。
你不需要考虑数组中超出新长度后面的元素。
说明:
为什么返回数值是整数，但输出的答案是数组呢?
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。
你可以想象内部操作如下:
// nums is passed in by reference. (i.e., without making a copy) int len = removeDuplicates(nums); // 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
使用快慢指针来记录遍历的坐标。
- 开始时这两个指针都指向第一个数字
- 如果两个指针指的数字相同，则快指针向前走一步
- 如果不同，则两个指针都向前走一步
- 当快指针走完整个数组后，慢指针当前的坐标加1就是数组中不同数字的个数
关键点解析
- 双指针
这道题如果不要求，O(n)的时间复杂度， O(1)的空间复杂度的话，会很简单。但是这道题是要求的，这种题的思路一般都是采用双指针
- 如果是数据是无序的，就不可以用这种方式了，从这里也可以看出排序在算法中的基础性和重要性。
代码
/* * @lc app=leetcode id=26 lang=javascript * * [26] Remove Duplicates from Sorted Array * * https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/ * * algorithms * Easy (39.76%) * Total Accepted: 539.7K * Total Submissions: 1.4M * Testcase Example: '[1,1,2]' * * Given a sorted array nums, remove the duplicates in-place such that each * element appear only once and return the new length. * * Do not allocate extra space for another array, you must do this by modifying * the input array in-place with O(1) extra memory. * * Example 1: * * * Given nums = [1,1,2], * * Your function should return length = 2, with the first two elements of nums * being 1 and 2 respectively. * * It doesn't matter what you leave beyond the returned length. * * Example 2: * * * Given nums = [0,0,1,1,1,2,2,3,3,4], * * Your function should return length = 5, with the first five elements of nums * being modified to 0, 1, 2, 3, and 4 respectively. * * It doesn't matter what values are set beyond the returned length. * * * Clarification: * * Confused why the returned value is an integer but your answer is an array? * * 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. * * Internally you can think of this: * * * // nums is passed in by reference. (i.e., without making a copy) * int len = removeDuplicates(nums); * * // 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]); * } * */ /** * @param {number[]} nums * @return {number} */ var removeDuplicates = function(nums) { const size = nums.length; let slowP = 0; for (let fastP = 0; fastP < size; fastP++) { if (nums[fastP] !== nums[slowP]) { slowP++; nums[slowP] = nums[fastP] } } return slowP + 1; };
