Merge Sorted Array

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 nums1and nums2 are m and n respectively.

问题：

两个有序数组，合并成一个有序数组，如果第一个数组空间足够容纳两个数组。

分析：

考虑到a数组非常大，能够直接在a数组上进行合并，可是要讲究效率。假设单纯从前往后合并。那么效率会非常低，由于a数组后面的数字须要不停的移动。换一种思路，我们採用从后往前合并，首先计算出总长度，设置一个指针从a数组最后往前移动。

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

#define MAX 1024

void combine(int *a, int *b, int len1, int len2)
{
    if(a == NULL || b == NULL || (len1 + len2) > MAX)
        return ;

    int new_point;
    int a_point = len1 - 1;
    int b_point = len2 - 1;

    new_point = len1 + len2 -1; //总的长度

    while(a_point >= 0 && b_point >= 0)
    {
        if(a[a_point] > b[b_point])
        {
            a[new_point--] = a[a_point--];
        }
        else
        {
            a[new_point--] = b[b_point--];
        }
    }

    while(a_point >= 0)
    {
        a[new_point--] = a[a_point--];
    }

    while(b_point >= 0)
    {
        a[new_point--] = b[b_point--];
    }

    return ;
}

int main()
{
    int b[MAX] = {1,2,3,4};
    int a[MAX] = {5,6,7,8};

    combine(a, b, 4, 4);

    for(int i =0 ; i <= 4 + 4 -1; i++)
    {
        cout << a[i] << " ";
    }

    return 0;
}

C++运用STL风格：

class Solution
{
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
    {
        vector<int> result;
        vector<int>::iterator iter1=nums1.begin();
        vector<int>::iterator iter2=nums2.begin();
        while(iter1!=nums1.begin()+m && iter2!=nums2.begin()+n)
        {
            if(*iter1<=*iter2)
            {
                result.push_back(*iter1);
                ++iter1;
            }
            else
            {
                result.push_back(*iter2);
                ++iter2;
            }
        }
        while(iter1!=nums1.begin()+m)
        {
            result.push_back(*iter1);
            ++iter1;
        }
        while(iter2!=nums2.begin()+n)
        {
            result.push_back(*iter2);
            ++iter2;
        }
        nums1.swap(result);
    }
};