在一个数组中实现两个栈，当数组未填满是任一个栈不能溢出。解法是将一个栈从头开始往后插入，而另一个从后往前插入，如果插入一个元素后，两个栈的top指针未相遇，则表示数组未满，，栈没有溢出。
;struct special_stack{int capcity;int ltop,rtop;int* vals;};special_stack* create(int capacity){special_stack* sstack = new special_stack;sstack->vals = new int[capacity];sstack->capcity = capacity;sstack->ltop = -1;sstack->rtop = capacity;return sstack;}bool lpush(special_stack** sstack,int val){if((*sstack)->ltop + 1 == (*sstack)->rtop) return false;(*sstack)->vals[++((*sstack)->ltop)] = val;return true;}bool lpop(special_stack** sstack,int& val){if((*sstack)->ltop < 0) return false;val = (*sstack)->vals[((*sstack)->ltop)–];return true;}bool rpush(special_stack** sstack,int val){if((*sstack)->rtop – 1 == (*sstack)->ltop) return false;(*sstack)->vals[–((*sstack)->rtop)] = val;return true;}bool rpop(special_stack** sstack,int& val){if((*sstack)->rtop >= (*sstack)->capcity) return false;val = (*sstack)->vals[((*sstack)->rtop)++];return true;}void destroy(special_stack** sstack){delete [](*sstack)->vals;delete (*sstack);*sstack = NULL;}void clear(special_stack** sstack){(*sstack)->ltop = -1;(*sstack)->rtop = (*sstack)->capcity;}int _tmain(int argc, _TCHAR* argv[]){special_stack* stack = create(20);for (int i = 1; i <= 10; ++i){lpush(&stack,i);rpush(&stack,i + 10);}for (int i = 1; i <= 10; ++i){int val;if(lpop(&stack,val))cout<<val<<‘ ‘;}cout<<endl;for (int i = 1; i <= 10; ++i){int val;if(rpop(&stack,val))cout<<val<<‘ ‘;}cout<<endl;for (int i = 1; i <= 10; ++i){lpush(&stack,i);rpush(&stack,i + 10);}cout<<lpush(&stack,21)<<endl;cout<<rpush(&stack,31)<<endl;int val;lpop(&stack,val);rpop(&stack,val);cout<<lpush(&stack,21)<<endl;cout<<rpush(&stack,31)<<endl;destroy(&stack);return 0;}
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