leetcode_题解_Merge Sorted Array _简单题
2014-11-11 22:14
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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 m andn respectively.
class Solution {
public:
void merge(int A[], int m, int B[], int n) {
for(int i=0; i<m; i++)
A[m-1+n-i]=A[m-1-i];//将A的元素整体后移n位
int ia=n,ib=0,cur=0;
for(int i=0; i<m+n; i++)
{
if(ia>=m+n)
A[cur++]=B[ib++];//A已经被访问完所有元素<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">*******容易忽略的情况</span>
else if(ib>=n)
A[cur++]=A[ia++];//B已经被访问完所有元素<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">*******容易忽略的情况</span>
else if(A[ia]<=B[ib])
A[cur++]=A[ia++];
else
A[cur++]=B[ib++];
}
}
};
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 andn respectively.
class Solution {
public:
void merge(int A[], int m, int B[], int n) {
for(int i=0; i<m; i++)
A[m-1+n-i]=A[m-1-i];//将A的元素整体后移n位
int ia=n,ib=0,cur=0;
for(int i=0; i<m+n; i++)
{
if(ia>=m+n)
A[cur++]=B[ib++];//A已经被访问完所有元素<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">*******容易忽略的情况</span>
else if(ib>=n)
A[cur++]=A[ia++];//B已经被访问完所有元素<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">*******容易忽略的情况</span>
else if(A[ia]<=B[ib])
A[cur++]=A[ia++];
else
A[cur++]=B[ib++];
}
}
};
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