Description:

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

My code:

/**
 * @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 cur1 = m - 1, cur2 = n - 1, curpos = m + n - 1; // cur1: nums1當前排序的元素, cur2: nums2當前排序的元素, curpos: nums1當前排序的下標
   while(cur1 > -1 && cur2 > -1) {
       if(nums1[cur1] > nums2[cur2]) {
           nums1[curpos] = nums1[cur1];
           curpos--;
           cur1--;
       } else {
           nums1[curpos] = nums2[cur2];
           curpos--;
           cur2--;
       }
   }
    while(cur2 > -1) {
        nums1[curpos] = nums2[cur2];
        curpos--;
        cur2--;
    }
};

Note:
思路不完全是自己想的，剛開始想得不對，想了很久都沒想出解法，最后參考了discuss里的思路，捋了一遍。
nums1的長度為m+n，從最后往前排，即從m+n-1排最大的，到nums1[0]