Merge Sorted Array

Given two sorted interger 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 nums.The number of elements initialized in nums1 and nums2 are m and n respectively.

按照归并排序的惯性思路，因为归并排序给定的是一个数组的两个区间，所以通常情况下会借助O(n)大小的辅助空间。思路如下：

class Solution
{
public:
    void merge(vector<int> &nums1, int m, vector<int> &nums2, int n)
    {
         vector<int> A;
         A.resize(m+n);
         int pa = 0, pb = 0, i = 0;
         while(pa < m && pb < n)
         {
               if(nums[pa] < nums1[pb])
                   A[i ++] = nums1[pa ++];
               else
                   A[i ++] = nums2[pb ++];
         }  
         while(pa < m)
                 A[i ++] = nums1[pa ++];
         while(pb < n)
                 A[i ++] = nums2[pb ++];
         nums1.resize(m+n);
         while(--i >= 0)
                 nums1[i] = A[i];
    }  
}            

　　但是上述题目给定的条件更加灵活，因此可以采用尾插法，这样就不再需要辅助空间的开销，理解上也更加简洁：

class Solution
{
public:
    void merge(vector<int> &nums1, int m, vector<int> &nums2, int n)
    {
        int pa = m-1, pb = n-1, i = m+n-1;
        while(pa >= 0 && pb >= 0)
        {
            if(nums1[pa] > nums2[pb])
                nums1[i --] = nums1[pa --];
            else
                nums1[i --] = nums2[pb --];
        }
        while(pb >= 0)
                nums1[i --] = nums[pb --];
    }   
}