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算法思想:首先,按顺序不断取下两个两个顺序表较小的结点存入新的顺序表中。然后,哪个表有剩余,就将剩下的部分加到新的顺序表后面。
书中答案使用了3个循环,我将他合并成了一个循环来实现,伪代码如下:
int Combin_List(SqList L1,SqList L2,SqList*L) { if (L1.Length + L2.Length >
MAXSIZE) { return ERROR; } int k = 0, k1 = 0, k2 = 0; while (1) { if
(L1.elems[k1] <= L2.elems[k2]&&k1<L1.Length) { L->elems[k++] = L1.elems[k1++];
L->Length++; if (k1 == L1.Length) { L1.elems[k1]= L2.elems[L2.Length-1];
//将顺序表超出的第一个值设为另一个表的最大值,从而保证另一个if语句正常执行至拷贝结束 } } if (L1.elems[k1] >=
L2.elems[k2] && k2 < L2.Length) { L->elems[k++] = L2.elems[k2++]; L->Length++;
if (k2 == L2.Length) { L2.elems[k2] = L1.elems[L1.Length - 1]; } } if
(L->Length == L1.Length + L2.Length) { return OK; } } }