leeocde-Merge Sorted Array

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

Note:

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

Example:

Input: nums1 = [1,2,3,0,0,0], m = 3 nums2 = [2,5,6], n = 3

Output: [1,2,2,3,5,6]

This is a more basic merge sorting problem. We know that the idea of ​​merging and sorting is divide and conquer. Then this question is "treatment". E.g:

To merge the two ordered subsequences [4,5,7,8] and [1,2,3,6] into the final sequence [1,2,3,4,5,6,7,8] ], look at the implementation steps.

In this question, we merge forward from the back of the array:

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int size = m + n;
	//use m and n since they are local to us
        m--;
        n--;
	//iterate the nums1 list backwards
        for(int i = size -1; i >=0; --i)
        {
	//if last number in nums1 vector is greater than last number is nums2 vector
	//we put that at current location pointed by i
            if(m >= 0 && n>= 0 && nums1[m] > nums2[n])
                nums1[i] = nums1[m--];
	//otherwise if we still have anything in nums2 array, it must be bigger or equal to what is in nums1 array so use that
            else if(m >= 0 && n>= 0 && nums1[m] < nums2[n])
                nums1[i] = nums2[n--];
    //the rest of nums1
            else if(m >= 0)
                nums1[i] = nums1[m--];
    //the rest of nums2
            else if(n >= 0)
                nums1[i] = nums2[n--];
        }
    }

};