- 括号匹配
- 前言
- 例如
- 算法思想
- 栈的定义及其操作
- 括号匹配核心代码
- 完整代码
- 运行测试
括号匹配问题是较为经典的栈的应用了,在Leetcode和数据结构书中时常出现
例如给定一个字符串,其包含( , ) ,{ , } , [ , ]以及数字,要求判断其括号是否匹配,即:
- 左括号必须用相同类型的右括号闭合
- 左括号必须以正确的顺序闭合
实例 1:
输入: "[2 + (1 - 3)] * 4" 输出:1
实例 2:
输入:"( ) )" 输出:0算法思想
- 当遇到 ‘(’ ,‘[’ ,‘{’ 时进行压栈操作,将其压入栈中
- 当遇到 ‘)’ ,‘]’, ‘}’ 时进行弹栈操作,若此时弹出的元素和此时遇到的括号匹配,则继续进行,否则直接退出(即不匹配)
- 为了防止出现单独的 ‘)’ ,’]’ 或 ‘}’ 导致判断错误,在栈的栈底添加了 ‘#’ 标识符,若匹配结束而最后弹出的不是 ‘#’ 则表示有多余单独的 ‘)’ ,’]’ 或 ‘}’ ,则直接退出(即也不匹配)
typedef struct CharStack
{
int top;
int data[MAXSIZE];
} *CharStackPtr,CharStack;
//输出栈
void outPutStack(CharStackPtr tempStack)
{
for(int i = 0; i <= tempStack->top; ++i)
{
printf("%c ",tempStack->data[i]);
}
printf("n");
}
//初始化栈
CharStackPtr CharStackInit()
{
CharStackPtr resultPtr = (CharStackPtr)malloc(sizeof(CharStack));
resultPtr->top = -1;
return resultPtr;
}
//压栈
void push(CharStackPtr tempStackPtr, char tempValue)
{
if(tempStackPtr->top >= MAXSIZE - 1)
{
printf("无法压入元素%c:堆栈已满!n",tempValue);
return ;
}
tempStackPtr->top ++;
tempStackPtr->data[tempStackPtr->top] = tempValue;
}
//弹栈
char pop(CharStackPtr tempStackPtr)
{
if (tempStackPtr->top == -1)
{
printf("不能弹出元素:堆栈为空!n");
return NULL;
}
tempStackPtr->top --;
return tempStackPtr->data[tempStackPtr->top+1];
}
括号匹配核心代码
//括号匹配
bool parenthesisMatching(char* tempString)
{
CharStackPtr tempStack = CharStackInit();
push(tempStack,'#');
char tempChar,tempPopedChar;
for(int i = 0; i< strlen(tempString); ++i)
{
tempChar = tempString[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 false;
}
return true;
}
完整代码
#include运行测试#include #include #define MAXSIZE 10 typedef struct CharStack { int top; int data[MAXSIZE]; } *CharStackPtr,CharStack; //输出栈 void outPutStack(CharStackPtr tempStack) { for(int i = 0; i <= tempStack->top; ++i) { printf("%c ",tempStack->data[i]); } printf("n"); } //初始化栈 CharStackPtr CharStackInit() { CharStackPtr resultPtr = (CharStackPtr)malloc(sizeof(CharStack)); resultPtr->top = -1; return resultPtr; } //压栈 void push(CharStackPtr tempStackPtr, char tempValue) { if(tempStackPtr->top >= MAXSIZE - 1) { printf("无法压入元素%c:堆栈已满!n",tempValue); return ; } tempStackPtr->top ++; tempStackPtr->data[tempStackPtr->top] = tempValue; } //弹栈 char pop(CharStackPtr tempStackPtr) { if (tempStackPtr->top == -1) { printf("不能弹出元素:堆栈为空!n"); return NULL; } tempStackPtr->top --; return tempStackPtr->data[tempStackPtr->top+1]; } //括号匹配 bool parenthesisMatching(char* tempString) { CharStackPtr tempStack = CharStackInit(); push(tempStack,'#'); char tempChar,tempPopedChar; for(int i = 0; i< strlen(tempString); ++i) { tempChar = tempString[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 false; } return true; } //测试 void parenthesisMatchingTest() { char *tempExpression; tempExpression = "[2 + (1 - 3)] * 4"; bool tempMatch = parenthesisMatching(tempExpression); printf("表达式%s括号匹配吗? %dn",tempExpression,tempMatch); tempExpression = "( ) )"; tempMatch = parenthesisMatching(tempExpression); printf("表达式%s括号匹配吗? %dn",tempExpression,tempMatch); tempExpression = "()()(())"; tempMatch = parenthesisMatching(tempExpression); printf("表达式%s括号匹配吗? %dn",tempExpression,tempMatch); tempExpression = "({}[])"; tempMatch = parenthesisMatching(tempExpression); printf("表达式%s括号匹配吗? %dn",tempExpression,tempMatch); tempExpression = ")("; tempMatch = parenthesisMatching(tempExpression); printf("表达式%s括号匹配吗? %dn",tempExpression,tempMatch); } int main() { parenthesisMatchingTest(); }
表达式[2 + (1 - 3)] * 4括号匹配吗? 1
表达式( ) )括号匹配吗? 0
表达式()()(())括号匹配吗? 1
表达式({}[])括号匹配吗? 1
表达式)(括号匹配吗? 0



