0026.RemoveDuplicatesfromSortedArray.js

// 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).
//  

// Constraints:

// 0 <= nums.length <= 3 * 104
// -100 <= nums[i] <= 100
// nums is sorted in non-decreasing order.

// 来源：力扣（LeetCode）
// 链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array
// 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

// 🎨 方法一：双指针

// 📝 思路：初始时first和last指针分别指向第一个和第二个元素，last向右遍历中，如果和first所在元素不通，则追加到first后面（替换），遍历结束，first的长度就是去重后数组的长度

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function (nums) {
    let len = nums.length

    if (len === 0) {
        return nums;
    }

    // 去重后数组末尾下标
    let first = 0;
    // 扫描数组
    let last = 1;

    while (last < len) {
        if (nums[last] === nums[first]) {
            last++
        } else {
            nums[++first] = nums[last++] //🔥
        }
    }

    nums.length = first + 1 //🔥
};