Merge Sorted Array II
最后更新于：2022-04-02 01:08:37
# Merge Sorted Array II
### Source
- lintcode: [(64) Merge Sorted Array II](http://www.lintcode.com/en/problem/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 mergeSortedArray(vector &A, vector &B) {
if (A.empty()) return B;
if (B.empty()) return A;
int aLen = A.size(), bLen = B.size();
vector 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 mergeSortedArray(ArrayList A, ArrayList B) {
if (A == null || A.isEmpty()) return B;
if (B == null || B.isEmpty()) return A;
ArrayList C = new ArrayList();
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)O(n)O(n), 空间复杂度 O(1)O(1)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。
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