【LeetCode】26. Remove Duplicates from Sorted Array 解题报告(Python&C++&Java)

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

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.

题目大意

从一个有序数组中删除重复的数字，只保留下无重复的有序数组，把这些数字放到原数组的前面部分，返回这部分的长度。

解题方法

双指针

慢指针指向应该放入元素的位置，每次移动一格。快指针找到应该放哪个元素，每次找到下一个新的元素。

Python代码如下：

class Solution(object):

    def removeDuplicates(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        N = len(nums)

        if N <= 1: return N

        left, right = 0, 1

        while right < N:

            while right < N and nums[right] == nums[left]:

                right += 1

            left += 1

            if right < N:

                nums[left] = nums[right]

        return left

C++代码如下：

class Solution {

public:

    int removeDuplicates(vector<int>& nums) {

        const int L = nums.size();

        if (L <= 1) return L;

        int slow = 1;

        int fast = 1;

        while (fast < L) {

            while (fast < L && nums[fast] == nums[fast - 1]) {

                fast ++;

            }

            if (fast < L) {

                nums[slow] = nums[fast];

                slow ++;

                fast ++;

            }

        }

        return slow;

    }

};

Java代码如下：

public int removeDuplicates(int[] nums) {

    if (nums.length == 0) return 0;

    int i = 0;

    for (int j = 1; j < nums.length; j++) {

        if (nums[j] != nums[i]) {

            i++;

            nums[i] = nums[j];

        }

    }

    return i + 1;

}

日期

2016 年 05月 8日
2019 年 9 月 17 日 —— 听了hulu宣讲会，觉得hulu的压力不大

巴特西

【LeetCode】26. Remove Duplicates from Sorted Array 解题报告(Python&C++&Java)

目录

题目描述

题目大意

解题方法

双指针

日期

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