我们把类似于弹夹那种先进后出的数据结构称为栈，栈是限定仅在表尾进行插入和删除操作的线性表，我们把允许插入和删除的一端称为栈顶，另一端称为栈底，不含任何数据元素的栈称为空栈，栈又称后进后出的线性表，简称LIFO结构。

栈首先是一个线性表，也就是说，栈元素具有线性关系，即前驱后继关系，只不过它是一种特殊的线性表而已。

栈的特殊之处在于限制了这个线性表的插入和删除位置，它始终只在栈顶进行。这也就使得：栈底是固定的，最先进栈的只能在栈底。

栈的插入操作，叫做进栈；栈的删除操作叫做出栈。

​
#include 
#include 

#define STACK_MAX_SIZE 10

typedef struct CharStack {
    int top;

    int data[STACK_MAX_SIZE]; 
} *CharStackPtr;

CharStackPtr charStackInit() {
	CharStackPtr resultPtr = (CharStackPtr)malloc(sizeof(CharStack));
	resultPtr->top = -1;

	return resultPtr;
}


void outputStack(CharStackPtr paraStack) {
    for (int i = 0; i <= paraStack->top; i ++) {
        printf("%c ", paraStack->data[i]);
    }
    printf("rn");
}

void push(CharStackPtr paraStackPtr, int paraValue) {
    if (paraStackPtr->top >= STACK_MAX_SIZE - 1) {
        printf("Cannot push element: stack full.rn");
        return;
    }

	paraStackPtr->top ++;

    paraStackPtr->data[paraStackPtr->top] = paraValue;
}

char pop(CharStackPtr paraStackPtr) {
    if (paraStackPtr->top < 0) {
        printf("Cannot pop element: stack empty.rn");
        return '';
    }

	paraStackPtr->top --;

    return paraStackPtr->data[paraStackPtr->top + 1];
}

void pushPopTest() {
    printf("---- pushPopTest begins. ----rn");

    CharStackPtr tempStack = charStackInit();
    printf("After initialization, the stack is: ");
	outputStack(tempStack);

	for (char ch = 'a'; ch < 'm'; ch ++) {
		printf("Pushing %c.rn", ch);
		push(tempStack, ch);
		outputStack(tempStack);
	}

	for (int i = 0; i < 3; i ++) {
		ch = pop(tempStack);
		printf("Pop %c.rn", ch);
		outputStack(tempStack);
	}

    printf("---- pushPopTest ends. ----rn");
}


void main() {
	pushPopTest();
}


​

是

括号匹配问题

#include 
#include 

#define STACK_MAX_SIZE 10

typedef struct CharStack {
    int top;

    int data[STACK_MAX_SIZE]; 
} *CharStackPtr;

CharStackPtr charStackInit() {
	CharStackPtr resultPtr = (CharStackPtr)malloc(sizeof(struct CharStack));
	resultPtr->top = -1;

	return resultPtr;
}

void outputStack(CharStackPtr paraStack) {
    for (int i = 0; i <= paraStack->top; i ++) {
        printf("%c ", paraStack->data[i]);
    }
    printf("rn");
}

void push(CharStackPtr paraStackPtr, int paraValue) {
    if (paraStackPtr->top >= STACK_MAX_SIZE - 1) {
        printf("Cannot push element: stack full.rn");
        return;
    }

	paraStackPtr->top ++;

    paraStackPtr->data[paraStackPtr->top] = paraValue;
}

char pop(CharStackPtr paraStackPtr) {
    if (paraStackPtr->top < 0) {
        printf("Cannot pop element: stack empty.rn");
        return '';
    }

	paraStackPtr->top --;

    return paraStackPtr->data[paraStackPtr->top + 1];
}

void pushPopTest() {
    printf("---- pushPopTest begins. ----rn");

    CharStackPtr tempStack = charStackInit();
    printf("After initialization, the stack is: ");
	outputStack(tempStack);

	for (char ch = 'a'; ch < 'm'; ch ++) {
		printf("Pushing %c.rn", ch);
		push(tempStack, ch);
		outputStack(tempStack);
	}

	for (int i = 0; i < 3; i ++) {
		ch = pop(tempStack);
		printf("Pop %c.rn", ch);
		outputStack(tempStack);
	}

    printf("---- pushPopTest ends. ----rn");
}

bool bracketMatching(char* paraString, int paraLength) {
    CharStackPtr tempStack = charStackInit();
	push(tempStack, '#');
	char tempChar, tempPopedChar;

	for (int i = 0; i < paraLength; i++) {
		tempChar = paraString[i];

		switch (tempChar) {
		case '(':
		case '[':
		case '{':
			push(tempStack, tempChar);
			break;
		case ')':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '(') {
				return false;
			}
			break;
		case ']':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '[') {
				return false;
			}
			break;
		case '}':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '{') {
				return false;
			} 
			break;
		default:
			break;
		}
	}

	tempPopedChar = pop(tempStack);
	if (tempPopedChar != '#') {
		return true;
	}

	return true;
}

void bracketMatchingTest() {
	char* tempExpression = "[2 + (1 - 3)] * 4";
	bool tempMatch = bracketMatching(tempExpression, 17);
	printf("Is the expression '%s' bracket matching? %d rn", tempExpression, tempMatch);


	tempExpression = "( )  )";
	tempMatch = bracketMatching(tempExpression, 6);
	printf("Is the expression '%s' bracket matching? %d rn", tempExpression, tempMatch);

	tempExpression = "()()(())";
	tempMatch = bracketMatching(tempExpression, 8);
	printf("Is the expression '%s' bracket matching? %d rn", tempExpression, tempMatch);

	tempExpression = "({}[])";
	tempMatch = bracketMatching(tempExpression, 6);
	printf("Is the expression '%s' bracket matching? %d rn", tempExpression, tempMatch);


	tempExpression = ")(";
	tempMatch = bracketMatching(tempExpression, 2);
	printf("Is the expression '%s' bracket matching? %d rn", tempExpression, tempMatch);
}

void main() {
	// pushPopTest();
	bracketMatchingTest();
}

表达式求值

#include 
#include 
#include 
#include 
#include 
 
using namespace std;
 
stack num;
stack op;
 
void eval()
{
    auto b = num.top();
    num.pop();
    auto a = num.top();
    num.pop();
    auto c = op.top();
    op.pop();
    int x;
    if (c == '+') x = a + b;
    else if (c == '-') x = a - b;
    else if (c == '*') x = a * b;
    else x = a / b;
    num.push(x);
}
 
int main()
{
    unordered_map pr{{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};
    string str;
    cin >> str;
    for (int i = 0; i < str.size(); i ++ )
    {
        auto c = str[i];
        if (isdigit(c))
        {
            int x = 0, j = i;
            while (j < str.size() && isdigit(str[j]))
                x = x * 10 + str[j ++ ] - '0';
            i = j - 1;
            num.push(x);
        }
        else if (c == '(') op.push(c);
        else if (c == ')')
        {
            while (op.top() != '(') eval();
            op.pop();
        }
        else
        {
            while (op.size() && op.top() != '(' && pr[op.top()] >= pr[c]) eval();
            op.push(c);
        }
    }
    while (op.size()) eval();
    cout << num.top() << endl;
    return 0;
}