数据结构-括号匹配

这是栈的运用（但是代码只实现了对数学式子的是否是正确形式的判断，没有实现计算功能）

1.闵帆老师的代码理解和分析

#include <stdbool.h> //bool函数的头文件 
#include <stdio.h>
#include <malloc.h>

#define STACK_MAX_SIZE 10

//建立栈 
typedef struct CharStack {
    int top;

    char data[STACK_MAX_SIZE]; //The maximum length is fixed.
} *CharStackPtr;

//输出data数组里面的所有值 
void outputStack(CharStackPtr paraStack) {
    for (int i = 0; i <= paraStack->top; i ++) {
        printf("%c ", paraStack->data[i]);
    }// Of for i
    printf("\r\n");
}// Of outputStack

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

	return resultPtr;
}//Of charStackInit

//进栈（data数组赋值） 
void push(CharStackPtr paraStackPtr, int paraValue) {
    // Step 1. Space check.
    if (paraStackPtr->top >= STACK_MAX_SIZE - 1) {
        printf("Cannot push element: stack full.\r\n");
        return;
    }//Of if

    // Step 2. Update the top.
	paraStackPtr->top ++;

	// Step 3. Push element.
    paraStackPtr->data[paraStackPtr->top] = paraValue;
}// Of push

//出栈 
char pop(CharStackPtr paraStackPtr) {
    // Step 1. Space check.
    if (paraStackPtr->top < 0) {
        printf("Cannot pop element: stack empty.\r\n");
        return '\0';
    }//Of if

    // Step 2. Update the top.
	paraStackPtr->top --;

	// Step 3. Pop element.
    return paraStackPtr->data[paraStackPtr->top + 1];
}// Of pop

/**
 * Test the push function.
 */
void pushPopTest() {
    printf("---- pushPopTest begins. ----\r\n");
    char ch;
    
	// Initialize.
    CharStackPtr tempStack = charStackInit();
    printf("After initialization, the stack is: ");
	outputStack(tempStack);

	// Pop.
	for (ch = 'a'; ch < 'm'; ch ++) {
		printf("Pushing %c.\r\n", ch);
		push(tempStack, ch);
		outputStack(tempStack);
	}//Of for i

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

    printf("---- pushPopTest ends. ----\r\n");
}// Of pushPopTest

//对一个数学式子，判断这个数值括号是否匹配 （核心） 
bool bracketMatching(char* paraString, int paraLength) {
	// Step 1. Initialize the stack through pushing a '#' at the bottom.
    CharStackPtr tempStack = charStackInit(); 
	push(tempStack, '#');
	char tempChar, tempPopedChar;

	// Step 2. Process the string.
	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;
			} // Of if
			break;
		case ']':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '[') {
				return false;
			} // Of if
			break;
		case '}':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '{') {
				return false;
			} // Of if
			break;
		default:
			// Do nothing.
			break;
		}// Of switch
	} // Of for i

	tempPopedChar = pop(tempStack); //出栈的符号 
	if (tempPopedChar != '#') {
		return false;
	} // Of if

	return true;
}// Of bracketMatching

/**
 * Unit test.
 */
void bracketMatchingTest() {
	char* tempExpression = "[2 + (1 - 3)] * 4";
	bool tempMatch = bracketMatching(tempExpression, 17);//17包括空格 
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);


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

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

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


	tempExpression = ")(";
	tempMatch = bracketMatching(tempExpression, 2);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);
}// Of bracketMatchingTest

/**
 The entrance.
 */
void main() {
	// pushPopTest();
	bracketMatchingTest();
}// Of main

2.对括号匹配知识的理解

1.将一个数学式子放到栈中

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


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

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

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


	tempExpression = ")(";
	tempMatch = bracketMatching(tempExpression, 2);
	printf("Is the expression '%s' bracket matching? %d \r\n", tempExpression, tempMatch);
}// Of bracketMatchingTest

bracketMatching（） 函数中输入paraLength，需要包括空格。

2.核心函数（括号匹配）

//对一个数学式子，判断这个数值括号是否匹配 （核心） 
bool bracketMatching(char* paraString, int paraLength) {
	// Step 1. Initialize the stack through pushing a '#' at the bottom.
    CharStackPtr tempStack = charStackInit(); 
	push(tempStack, '#');
	char tempChar, tempPopedChar;

	// Step 2. Process the string.
	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;
			} // Of if
			break;
		case ']':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '[') {
				return false;
			} // Of if
			break;
		case '}':
			tempPopedChar = pop(tempStack);
			if (tempPopedChar != '{') {
				return false;
			} // Of if
			break;
		default:
			// Do nothing.
			break;
		}// Of switch
	} // Of for i

	tempPopedChar = pop(tempStack); //出栈的符号 
	if (tempPopedChar != '#') {
		return false;
	} // Of if

	return true;
}// Of bracketMatching

 代码里是忽略了所有非括号字符，凭借括号的前后顺序判断式子是否是正确形式。

举例了代码中的情况1

在后续的【数据结构-表达式求值】的学习中，会学会代码，使其满足得到一个数学式子的值的要求。