26. Remove Duplicates from Sorted Array

Leetcode

https://leetcode.com/problems/remove-duplicates-from-sorted-array/

題目

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

解答

• 方法一 - Two pointer

仿照第 27. 題用雙指針的解法，先固定左指針，右指針從左指針往後找到 大於左指針數值的元素，再與左指針後的下一個元素進行交換

var removeDuplicates = function (nums) {
    let output = 1;

    let i = 0;
    let j = 1;

    const swap = (i, j) => {
        const temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
        output++;
    }

    while (i < nums.length - 1) {
        let value = nums[i];

        while (j < nums.length) {
            // 右指針往後找到大於 i 所在位置的元素
            if (nums[j] <= value) {
                j++;
            } else {
                // 交換元素
                swap(i + 1, j);
                break;
            }
        }
        // 左指針往後
        i++;
    }

    return output;
};

Runtime: 62 ms Beats 68.37% of users with JavaScript

Memory: 51.83 MB Beats 75.07% of users with JavaScript

• 方法二

利用 splice 方法，當遇到前一個元素重複時直接移除掉

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    for (let i=0; i<nums.length; i++) {
        if (nums[i] === nums[i+1]) {
            nums.splice(i, 1);
            i--;
        };
    }

    return nums.length;
};

方法三 - 快慢指針

由於矩陣已經排序過了，所以我們可以使用快慢指針的方式來將 非重複的值 與前面重複的值進行交換

範例：

nums = [0,0,1,1,1,2,2,3,3,4]

首先一開始固定 慢指針(slow) 在第一個元素的位置，快指針(fast) 在第二個元素的位置，快指針會不斷地往後移動，當遇到慢指針所在的元素與快指針所在的元素不同時，就可以將慢指針的下一個元素與快指針的元素交換，這樣就可以把快指針所在的 非重複的值 依序往前交換移動

var removeDuplicates = function (nums) {
    let slow = 0;

    for (let fast = 1; fast < nums.length; fast++) {
        if (nums[slow] !== nums[fast]) {
            slow++;
            nums[slow] = nums[fast];
        };
    }

    return slow + 1;
};

Runtime: 67 ms Beats 47.16% of users with JavaScript

Memory: 52.41 MB Beats 36.69% of users with JavaScript

測資

let nums = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4];
nums = [1, 1, 2];
nums = [1];

const result = removeDuplicates(nums)
console.log(nums, result);

參考資料

https://ithelp.ithome.com.tw/articles/10270237ithelp.ithome.com.tw