leetcode 88 Merge Sorted Array

 

 

Given two sorted integer arrays nums1 and nums2, merge nums2 intonums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal tom + n) to hold additional elements from nums2. The number of elements initialized innums1 and nums2 are m and n respectively.

 

測试用例：

 

Runtime Error Message: Last executed input: Input:[1,2,3,0,0,0], 3, [2,5,6], 3 Output:[1,2,3,5,6] Expected:[1,2,2,3,5,6]

 

错误的解决方式：

 

class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
{
set<int> result;
for(int i = 0;i<m;i++)
{
result.insert(nums1[i]);
}
for(int i = 0;i<n;i++)
{
result.insert(nums2[i]);
}
nums1.clear();
set<int>::iterator iter = result.begin();
for(;iter!=result.end();iter++)
{
nums1.push_back(*iter);
}
}
};

 

我的解决方式：上面就是同样 的元素没装进来，换成multiset即可了

class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
{
multiset<int> result;
for(int i = 0;i<m;i++)
{
result.insert(nums1[i]);
}
for(int i = 0;i<n;i++)
{
result.insert(nums2[i]);
}
nums1.clear();
set<int>::iterator iter = result.begin();
for(;iter!=result.end();iter++)
{
nums1.push_back(*iter);
}
}
};

 简短的解决方式：

class Solution {
public:
void merge(int A[], int m, int B[], int n) {
int k = m + n;
while (k-- > 0)
A[k] = (n == 0 || (m > 0 && A[m-1] > B[n-1])) ?  A[--m] : B[--n];
}
};

可读性较好：

class Solution {
public:
void merge(int A[], int m, int B[], int n) {
int i=m-1;
int j=n-1;
int k = m+n-1;
while(i >=0 && j>=0)
{
if(A[i] > B[j])
A[k--] = A[i--];
else
A[k--] = B[j--];
}
while(j>=0)
A[k--] = B[j--];
}
};

python解决方式：

class Solution:
# @param A  a list of integers
# @param m  an integer, length of A
# @param B  a list of integers
# @param n  an integer, length of B
# @return nothing(void)
def merge(self, A, m, B, n):
x=A[0:m]
y=B[0:n]
x.extend(y)
x.sort()
A[0:m+n]=x

python解决方式2：thats why we love python

def merge(self, A, m, B, n):
A[m:] = B[:n]
A.sort()