数据结构与算法——栈的应用括号匹配_数据结构括号匹配算法

  括号匹配问题是对于栈的应用，那么如何利用栈解决括号匹配的问题呢？

一、引入概念

一个表达式中利用圆括号或方括号及花括号来表示运算的优先级，将这些括号提取出来就构成了括号序列

例如：表达式[(a+b)*c]-[e-f] 其括号序列为[()][]

合法的括号序列称为匹配序列，不合法的括号序列称为不匹配序列

匹配序列示例： （[()]） [][]() ()[()]

不匹配序列示例：（[()] ][]() (][()]

二、算法分析

1.计算机从左往右扫描表达式，扫描至左括号时，将该左括号压入栈中。

2.直至遇见右括号，将栈中栈底的括号与该右括号进行比较，若结果为匹配，则继续进行扫描。若结果为不匹配，则该序列不合法。

3.若全部元素遍历完毕，栈中无剩余元素，则该序列合法；栈中仍然存在元素，则该序列不合法。

三、过程分析

1.初始化空栈，判断1[为左括号，将1入栈

2.判断2(，2仍为左括号，将2入栈

3.判断3），3为右括号，与栈顶元素（匹配，匹配成功，2出栈

4.判断4],4为右括号，与栈顶元素1[进行匹配，匹配成功，1出栈

5.判断5[,5为左括号，将5入栈

6.判断6]，6为右括号，与栈顶元素匹配，匹配成功，5出栈

7.序列遍历完毕，且栈中无元素，判断该序列为合法序列

（该案例及图片来自于知乎“半颗糖逗”）

四、代码如下

#include <stdio.h>
#include <malloc.h>
 
#define STACK_MAX_SIZE 10
 
/**
 * Linear stack of integers. The key is data.
 */
typedef struct CharStack {
    int top;
 
    int data[STACK_MAX_SIZE]; //The maximum length is fixed.
} *CharStackPtr;
 
/**
 * Output the stack.
 */
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
 
/**
 * Initialize an empty char stack. No error checking for this function.
 * @param paraStackPtr The pointer to the stack. It must be a pointer to change the stack.
 * @param paraValues An int array storing all elements.
 */
CharStackPtr charStackInit() {
	CharStackPtr resultPtr = (CharStackPtr)malloc(sizeof(struct CharStack));
	resultPtr->top = -1;
 
	return resultPtr;
}//Of charStackInit
 
/**
 * Push an element to the stack.
 * @param paraValue The value to be pushed.
 */
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
 
/**
 * Pop an element from the stack.
 * @return The popped value.
 */
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. Push 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
 
/**
 * Is the bracket matching?
 * 
 * @param paraString The given expression.
 * @return Match or not.
 */
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);
	if (tempPopedChar != '#') {
		return true;
	} // Of if
 
	return true;
}// Of bracketMatching
 
/**
 * Unit test.
 */
void bracketMatchingTest() {
	char* tempExpression = "[2 + (1 - 3)] * 4";
	bool tempMatch = bracketMatching(tempExpression, 17);
	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

(该代码誊抄自于闵帆老师)

五、运行结果

Is the expression '[2 + (1 - 3)] * 4' bracket matching? 1
Is the expression '( )  )' bracket matching? 0
Is the expression '()()(())' bracket matching? 1
Is the expression '({}[])' bracket matching? 1
Is the expression ')(' bracket matching? 0
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