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.

給定兩個有序整數數組 nums1和nums2，將nums2合併到 nums1 中，使得 num1成為一個有序數組。

<解題>

1. 把nums1和nums2的最後一個不含0字元設指標，接著比大小，比較大的從nums1的最後一位開始放，放完之後指標往前移
2. p2<0，代表 nums2 數組中的元素已經全部合併到 nums1，所以把num1剩下的元素處理完
3. p1>= 0，要確認num1的元素都有處理到

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        if (null != nums1 && null != nums2 && nums1.length == m + n) {
            int p1 = m - 1
            int p2 = n - 1;
            for (int i = m + n - 1; i >= 0; i--) {
                if (p2 < 0 ||(p1 >= 0 && nums1[p1] > nums2[p2])) {
                    nums1[i] = nums1[p1];
                    p1--;
                } else {
                    nums1[i] = nums2[p2];
                    p2--;
                }
            }
        }
    }
}