微软面试100题系列-设计包含 min 函数的栈

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定义栈的数据结构，要求添加一个 min 函数，能够得到栈的最小元素。

要求函数 min、push 以及 pop 的时间复杂度都是 O(1)。

#define STACK_SZ 10struct stack{int buf[STACK_SZ];int top_index;int min_index;};int is_empty(struct stack * s){return s->top_index == -1;}int is_full(struct stack * s){return s->top_index == STACK_SZ - 1;}int push(struct stack * s, int elem){if(is_full(s))return 0;s->top_index++;s->buf[s->top_index] = elem;if(s->min_index == -1)s->min_index = s->top_index;else if(s->buf[s->min_index] > elem)s->min_index = s->top_index;return 1;}int pop(struct stack * s, int * elem){if(is_empty(s))return 0;*elem = s->buf[s->top_index];s->top_index--;return 1;}void init_stack(struct stack * s){s->top_index = s->min_index = -1;}int get_min(struct stack * s, int * elem){if(s->min_index == -1)return 0;*elem = s->buf[s->min_index];return 1;}int main(int c, char * v[]){struct stack s;int min;init_stack(&s);if(get_min(&s, &min))printf("min value is : %d\n", min);push(&s, 4);push(&s, 3);push(&s, 4);push(&s, 4);if(get_min(&s, &min))printf("min value is : %d\n", min);}

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