Merge Sorted Array II

Source

• lintcode: (64) Merge Sorted Array II

Merge two given sorted integer array A and B into a new sorted integer array.

Example
A=[1,2,3,4]

B=[2,4,5,6]

return [1,2,2,3,4,4,5,6]

Challenge
How can you optimize your algorithm
if one array is very large and the other is very small?

题解

上题要求 in-place, 此题要求返回新数组。由于可以生成新数组，故使用常规思路按顺序遍历即可。

Python

class Solution:
    #@param A and B: sorted integer array A and B.
    #@return: A new sorted integer array
    def mergeSortedArray(self, A, B):
        if A is None or len(A) == 0:
            return B
        if B is None or len(B) == 0:
            return A

        C = []
        aLen, bLen = len(A), len(B)
        i, j = 0, 0
        while i < aLen and j < bLen:
            if A[i] < B[j]:
                C.append(A[i])
                i += 1
            else:
                C.append(B[j])
                j += 1

        # A has elements left
        while i < aLen:
            C.append(A[i])
            i += 1

        # B has elements left
        while j < bLen:
            C.append(B[j])
            j += 1

        return C

C++

class Solution {
public:
    /**
     * @param A and B: sorted integer array A and B.
     * @return: A new sorted integer array
     */
    vector<int> mergeSortedArray(vector<int> &A, vector<int> &B) {
        if (A.empty()) return B;
        if (B.empty()) return A;

        int aLen = A.size(), bLen = B.size();
        vector<int> C;
        int i = 0, j = 0;
        while (i < aLen && j < bLen) {
            if (A[i] < B[j]) {
                C.push_back(A[i]);
                ++i;
            } else {
                C.push_back(B[j]);
                ++j;
            }
        }

        // A has elements left
        while (i < aLen) {
            C.push_back(A[i]);
            ++i;
        }

        // B has elements left
        while (j < bLen) {
            C.push_back(B[j]);
            ++j;
        }

        return C;
    }
};

Java

class Solution {
    /**
     * @param A and B: sorted integer array A and B.
     * @return: A new sorted integer array
     */
    public ArrayList<Integer> mergeSortedArray(ArrayList<Integer> A, ArrayList<Integer> B) {
        if (A == null || A.isEmpty()) return B;
        if (B == null || B.isEmpty()) return A;

        ArrayList<Integer> C = new ArrayList<Integer>();
        int aLen = A.size(), bLen = B.size();
        int i = 0, j = 0;
        while (i < aLen && j < bLen) {
            if (A.get(i) < B.get(j)) {
                C.add(A.get(i));
                i++;
            } else {
                C.add(B.get(j));
                j++;
            }
        }

        // A has elements left
        while (i < aLen) {
            C.add(A.get(i));
            i++;
        }

        // B has elements left
        while (j < bLen) {
            C.add(B.get(j));
            j++;
        }

        return C;
    }
}

源码分析

分三步走，后面分别单独处理剩余的元素。

复杂度分析

遍历 A, B 数组各一次，时间复杂度 $O(n)$, 空间复杂度 $O(1)$.

Challenge

两个倒排列表，一个特别大，一个特别小，如何 Merge？此时应该考虑用一个二分法插入小的，即内存拷贝。