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Merge Sorted Array II.cpp
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Merge Sorted Array II.cpp
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Merge Sorted Array II
Source
lintcode: (64) Merge Sorted Array II
Given two sorted integer arrays A and B, merge B into A as one sorted array.
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 mand n respectively.
Example
A = [1, 2, 3, empty, empty] B = [4,5]
After merge, A will be filled as [1,2,3,4,5]
题解
在上题的基础上加入了in-place的限制。将上题的新数组视为length相对较大的数组即可,仍然从数组末尾进行归并,取出较大的元素。
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) {
int index = n + m;
while (m > 0 && n > 0) {
if (A[m - 1] > B[n - 1]) {
A[--index] = A[--m];
} else {
A[--index] = B[--n];
}
}
while (n > 0) {
A[--index] = B[--n];
}
while (m > 0) {
A[--index] = A[--m];
}
}
};
源码分析
因为本题有了 in-place 的限制,则必须从数组末尾的两个元素开始比较;否则就会产生挪动,一旦挪动就会是 O(n2) 的。
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 = n + m;
while (m > 0 && n > 0) {
if (A[m - 1] > B[n - 1]) {
A[--index] = A[--m];
} else {
A[--index] = B[--n];
}
}
while (n > 0) {
A[--index] = B[--n];
}
while (m > 0) {
A[--index] = A[--m];
}
}
};