1.已知两个有序数组A,B，将它们合并为一个有序数组。利用到的是归并算法的思想。
int* combine(int a[],int n1,int b[],int n2)
{
int i = 0,j = 0,k = 0;
int *c = new int[n1+n2];
while(i<n1&&j<n2) //依次比较a,b数组当前元素，将小的元素放前面，下标后移
{
if(a[i]<=b[j])
c[k++] = a[i++];
else
c[k++] = b[j++];
}
while(i<n1) //将剩余元素拷贝过去
{
c[k++] = a[i++];
}
while(j<n2)
{
c[k++] = b[j++];
}
return c;
}
2.删除数组中重复元素
void delete_same(int a[],int &n)
{
int i,j;
if(n==0)
return;
else
{
for(i = 0,j = 1;j<n;j++)
{
if(a[i]<a[j])
a[++i] = a[j];
}
}
n = i+1;
}
3.已知一个数组A，要求将数组元素循环左移p位。如一个数组A{a1,a2,a3,a4,a5}循环左移3位变成A{a4,a5,a1,a2,a3}.
一个思路是利用辅助数组，时间复杂度为O(n),空间复杂度为O(p)。还有一种高效的策略是将A的前P个元素逆置，再将后n-p个元素逆置，最后逆置整个数组。时间复杂度为O(n),空间复杂度为O(1)
void reverseA(int a[],int b,int e) //将b-e之间的数组逆序
{
int mid = (b+e)/2;
for(int i = b;i < mid;i++)
{
int t = a[i];
a[i] = a[e-(i-b)];
a[e-(i-b)] = t;
}
}void moveA(int a[],int n,int p)
{
reverseA(a, 0, p-1);
reverseA(a, p, n-1);
reverseA(a, 0, n-1);
}
4.两个有序序列，查找这两个有序序列的中位数。
int findmid(int a[],int b[],int n)
{
int s1 = 0,e1 = n-1;
int s2 = 0,e2 = n-1;
while(s1 != e1&&s2 != e2){
int m1 = (s1+e1)/2;
int m2 = (s2+e2)/2;
if(a[m1]==b[m2])
return a[m1];
else if(a[m1]<b[m2]){
if((s1+d1)%2==0)
{
s1 = m1;
e2 = m2;
}
else
{
s1 = m1+1;
e2 = m2;
}
}
else
{
if((s2+d2)%2==0)
{
e1 = m1;
s2 = m2;
}
else
{
e1 = m1;
s2 = m2+1;
}
}
}
return a[s1]<b[s2]?a[s1]:b[s2];
}
时间： 2024-12-31 06:18:21