0088. Merge Sorted Array

https://leetcode.com/problems/merge-sorted-array

Description

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

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

ac

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {

        int a=m-1;
        int b=n-1;
        int i=m+n-1;    // calculate the index of the last element of the merged array

        // go from the back by A and B and compare and put to the A element which is larger
        while(a>=0 && b>=0){
            if(A[a]>B[b])   A[i--]=A[a--];
            else            A[i--]=B[b--];
        }

        // if B is longer than A just copy the rest of B to A location, otherwise no need to do anything
        while(b>=0)         A[i--]=B[b--];
    }
};