Merge Sorted Array
Question
- leetcode: Merge Sorted Array | LeetCode OJ
- lintcode: (6) Merge Sorted Array
Given two sorted integer arrays A and B, merge B into A as one sorted array.
Example
A = [1, 2, 3, empty, empty], B = [4, 5]
After merge, A will be filled as [1, 2, 3, 4, 5]
Note
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.
題解
因為本題有 in-place 的限制,故必須從陣列末尾的兩個元素開始比較;否則就會產生挪動,一旦挪動就會是 的。
自尾部向首部逐個比較兩個陣列內的元素,取較大的置於陣列 A 中。由於 A 的容量較 B 大,故最後 m == 0
或者 n == 0
時僅需處理 B 中的元素,因為 A 中的元素已經在 A 中,無需處理。
Python
class Solution:
"""
@param A: sorted integer array A which has m elements,
but size of A is m+n
@param B: sorted integer array B which has n elements
@return: void
"""
def mergeSortedArray(self, A, m, B, n):
if B is None:
return A
index = m + n - 1
while m > 0 and n > 0:
if A[m - 1] > B[n - 1]:
A[index] = A[m - 1]
m -= 1
else:
A[index] = B[n - 1]
n -= 1
index -= 1
# B has elements left
while n > 0:
A[index] = B[n - 1]
n -= 1
index -= 1
C++
class Solution {
public:
/**
* @param A: sorted integer array A which has m elements,
* but size of A is m+n
* @param B: sorted integer array B which has n elements
* @return: void
*/
void mergeSortedArray(int A[], int m, int B[], int n) {
int index = m + n - 1;
while (m > 0 && n > 0) {
if (A[m - 1] > B[n - 1]) {
A[index] = A[m - 1];
--m;
} else {
A[index] = B[n - 1];
--n;
}
--index;
}
// B has elements left
while (n > 0) {
A[index] = B[n - 1];
--n;
--index;
}
}
};
Java
class Solution {
/**
* @param A: sorted integer array A which has m elements,
* but size of A is m+n
* @param B: sorted integer array B which has n elements
* @return: void
*/
public void mergeSortedArray(int[] A, int m, int[] B, int n) {
if (A == null || B == null) return;
int index = m + n - 1;
while (m > 0 && n > 0) {
if (A[m - 1] > B[n - 1]) {
A[index] = A[m - 1];
m--;
} else {
A[index] = B[n - 1];
n--;
}
index--;
}
// B has elements left
while (n > 0) {
A[index] = B[n - 1];
n--;
index--;
}
}
}
源碼分析
第一個 while 只能用條件與(conditional AND)。
複雜度分析
最壞情況下需要遍歷兩個陣列中所有元素,時間複雜度為 . 空間複雜度 .