[Algorithm-Elementary-Array] Delete duplicates in sorted array (multi-language version implementation)

[Algorithm-Elementary-Array] Delete duplicates in sorted array (multi-language version implementation)

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🍄 Algorithm Description

Language is only a means to implement algorithms, ideas are the most important.

If there are multiple solutions, choose only one language as the solution comparison.

If a certain algorithm is implemented alone, it will be implemented in multiple languages, and the characteristics of the languages will be compared.

😎Because if it is many-to-many, the space will be larger, which will affect the viewing experience!

🍖 topic

address

26. Remove duplicates in an ordered array

Given an ordered array nums, please delete the repeated elements in place so that each element appears only once, and return the new length of the deleted array.

Topic description

Do not use extra space for an array, you must modify the input array in place and in use under conditions of O (1) extra space to complete .

instruction

Why is the returned value an integer, but the output answer is an array?

Note that the input array is passed "by reference", which means that changes to the input array within the function are visible to the caller.

You can imagine the internal operation as follows:

// nums 是以“引用”方式传递的。也就是说，不对实参做任何拷贝
int len = removeDuplicates(nums);

// 在函数里修改输入数组对于调用者是可见的。
// 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。
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Example 1

输入：nums = [1,1,2]
输出：2, nums = [1,2]
解释：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 。不需要考虑数组中超出新长度后面的元素。

Example 2

输入：nums = [0,0,1,1,1,2,2,3,3,4]
输出：5, nums = [0,1,2,3,4]
解释：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 。不需要考虑数组中超出新长度后面的元素。

hint

0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums 已按升序排列

🥕 Ideas

violent thoughts

(Note that it is a violent idea, not a violent solution!)

As a straight man, I just want to achieve it.

Direct traversal, see that the title has been determined to be in order, and if it is not equal to the previous one, directly get it and store it in a new array. After traversing the direct new array is the answer.

It looks like it's very close! It's a simple subject after all.

but and under the condition of O(1) extra space. This is a hurdle that we cannot overcome. In this case, you have to consider operating on the original array.

Operate on the original array, first on the double pointer!

double pointer

ideas

The fast and slow pointers play, the fast pointer is fast , and the slow pointer is low .

The array is ordered, so repeated elements must be adjacent. Operate in the same array, that is, move the elements that are not repeated to the left of the array, and finally take the value of the array on the left.

Algorithm flow

• Compares the elements at positions fast and low

  • Loop execution:

    • If equal, fast shifted by 1 bit
    • If not equal, the low value before a changed fast , low after a shift, fast backward 1

  • End of loop:

    • fast bounds
• End of loop, return new array length low + 1

diagram

This will be the most soulful moment. Algorithms without diagrams are hooligans! (Hahaha, I will try to correct my previous hooliganism!)

In fact, it took me a lot of time to draw pictures, but I don't think it's a loss, I remember it more deeply!

Let's make a GIF! (Wrote a small tool to generate GIF from pictures in Python)

JavaScript

There is nothing special to say. If I force it, I will take the length calculation of nums

code

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    // 判断边界
    if (nums && nums.length < 0) {
        return 0
    }
    var low = 0, fast = 1, n = nums.length;
    while (fast < n) {
        if (nums[fast] !== nums[low]) {
          nums[low + 1] = nums[fast]
            low++  
        } 
        fast++
    }
    return low + 1
};

Java

Code

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length < 1) {
            return 0;
        }
        int fast = 1, low = 0, n = nums.length;
        while(fast < n) {
            if (nums[fast] != nums[low]){
                nums[low + 1] = nums[fast];
                low++;
            }
            fast++;
        }
        return low + 1;
    }
}

Python3

Code

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        if not nums:
            return 0
        n = len(nums)
        fast = 1
        low = 0
        while fast < n:
            if nums[fast] != nums[low]:
                nums[low + 1] = nums[fast]
                low += 1
            fast += 1
        return low + 1

PHP

Code

class Solution {

    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function removeDuplicates(&$nums) {
        if ($nums == null || count($nums) < 0) {
            return 0;
        }
        $fast = 1;
        $low = 0;
        $n = count($nums);
        while ($fast < $n) {
            if ($nums[$fast] != $nums[$low]) {
                $nums[$low + 1] = $nums[$fast];
                $low++;
            }
            $fast++;
        }
        return $low + 1;
    }
}

Go

Note that the go language has no while

code

func removeDuplicates(nums []int) int {
    n := len(nums)
    if n < 1 {
        return 0
    }
    low := 0
    for fast := 1; fast < n; fast++ {
        if nums[fast] != nums[low] {
            nums[low + 1] = nums[fast]
            low++
        }
    }
    return low + 1
}

C++

code

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = nums.size();
        if (n < 1) {
            return 0;
        }
        int fast = 1, low = 0;
        while (fast < n) {
            if (nums[fast] != nums[low]) {
                nums[low + 1] = nums[fast];
                low++;
            }
            fast++;
        }
        return low + 1;
    }
};

C

code

int removeDuplicates(int* nums, int numsSize){
    if(numsSize == 0) {
        return 0;
    }
    int fast = 1, low = 0;
    while (fast < numsSize) {
        if (nums[fast] != nums[low]) {
            nums[low + 1] = nums[fast];
            low++;
        }
        fast++;
    }
    return low + 1;
}

🥦 Summary

Start with arrays, think about how to extract ideas 🤔 array extraction, disassemble step by step from simple to complex, and also improve the data usage skills of programming languages.