标签：style c class blog code java
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]; }
标签：style c class blog code java
原文地址：http://www.cnblogs.com/cliviazhou/p/3753763.html
