【简单】Lintcode 64:Merge Sorted Array

最新推荐文章于 2020-09-28 09:12:21 发布

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本文介绍了一种合并两个已排序整数数组的方法，通过创建一个新的数组C来存储合并后的结果，并详细解释了具体的实现步骤。

Given two sorted integer arrays A and B, merge B into A as one sorted array.

Notice

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

Example

A = [1, 2, 3, empty, empty], B = [4, 5]

After merge, A will be filled as [1, 2, 3, 4, 5]

解题思路：

1、new一个大小为[m+n]的数组C，它的作用是用来暂时存放A与B的合并的结果。merge过程与归并排序的归并过程相似，大概就是用两个索引i,j分别指向A,B数组的下标，依次比较A[i]与B[j]的大小，将两者中更小的元素放进C[k]中，再移动相应索引到下一个比较元素...

class Solution {
public:
    /*
     * @param A: sorted integer array A which has m elements, but size of A is m+n
     * @param m: An integer
     * @param B: sorted integer array B which has n elements
     * @param n: An integer
     * @return: nothing
     */
    void mergeSortedArray(int A[], int m, int B[], int n)
    {
        // write your code here
	int * C = new int [m + n];

	int i = 0;
	int j = 0;
	int k = 0;
	
	while (k < m+n)
	{
		if (j >= n)
			C[k++] = A[i++];
		else if (i >= m)
			C[k++] = B[j++];

		else if (A[i] < B[j])
			C[k++] = A[i++];
		else
			C[k++] = B[j++];
	}
	
	//将C中的元素依次赋值给A
	for (int p = 0; p < (m + n); p++)
	{
		A[p] = C[p];
	}

	delete[] C;
    }
};