0088.MergeSortedArray.js

// You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

// Merge nums1 and nums2 into a single array sorted in non-decreasing order.

// The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

//  

// Example 1:

// Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
// Output: [1,2,2,3,5,6]
// Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
// The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
// Example 2:

// Input: nums1 = [1], m = 1, nums2 = [], n = 0
// Output: [1]
// Explanation: The arrays we are merging are [1] and [].
// The result of the merge is [1].
// Example 3:

// Input: nums1 = [0], m = 0, nums2 = [1], n = 1
// Output: [1]
// Explanation: The arrays we are merging are [] and [1].
// The result of the merge is [1].
// Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
//  

// Constraints:

// nums1.length == m + n
// nums2.length == n
// 0 <= m, n <= 200
// 1 <= m + n <= 200
// -109 <= nums1[i], nums2[j] <= 109
//  

// 来源：力扣（LeetCode）
// 链接：https://leetcode-cn.com/problems/merge-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：双指针分别逆序遍历两个数组，将较大值 unshift 到长数组末尾（赋值替换）

/**
 * @param {number[]} nums1
 * @param {number} m
 * @param {number[]} nums2
 * @param {number} n
 * @return {void} Do not return anything, modify nums1 in-place instead.
 */
var merge = function (nums1, m, nums2, n) {

    let p1 = m - 1;
    let p2 = n - 1;
    let p3 = m + n - 1;

    while (p1 >= 0 && p2 >= 0) {
        if (nums1[p1] > nums2[p2]) {
            nums1[p3] = nums1[p1];
            p1--;
            p3--;
        } else {
            nums1[p3] = nums2[p2];
            p2--;
            p3--;
        }
    }

    if (p2 >= 0) {
        for (let i = 0; i <= p2; i++) {
            nums1[i] = nums2[i]
        }
    }

    return nums1
};