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.

 

void merge(int* nums1, int m, int* nums2, int n) {
    int i = 0, j = 0, k = 0;
    int *tmp = malloc(m * sizeof(int));
    memcpy(tmp, nums1, m*sizeof(int));
    while (i < m && j < n)
    {
        if (tmp[i] <= nums2[j])
            nums1[k++] = tmp[i++];
        else
            nums1[k++] = nums2[j++];
    }
    while (j < n)
        nums1[k++] = nums2[j++];
    while (i < m)
        nums1[k++] = tmp[i++];
}

• 把nums1复制到另一个数组；然后把两个数组的内容复制到nums1中
• 注意：以变量为大小的数组需要使用malloc来开辟空间

class Solution {
public:
    void merge(int A[], int m, int B[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int index = m + n - 1;
        int aIndex = m - 1;
        int bIndex = n - 1;
        while(0 <= aIndex && 0 <= bIndex)
        {
            if (B[bIndex] > A[aIndex])
            {
                A[index--] = B[bIndex--];
            }
            else
            {
                A[index--] = A[aIndex--];
            }
        }
        
        while(0 <= aIndex)
        {
            A[index--] = A[aIndex--];
        }
        
        while(0 <= bIndex)
        {
            A[index--] = B[bIndex--];
        }
    }
};

• 这个方法好，从大到小来复制，这样就不用考虑位置被占用的问题