面试题整理12 求字符串括号最大深度子串

来源：互联网发布：淘宝打折软件叫什么编辑：程序博客网时间：2024/05/01 23:22

题目：求一个表达式字符串中括号层次最深的第一个表达式，如表达式

"a+b+((c+d)+(e+f))"，则结果为”c+d"。

分析：一般算式会让人想起用栈分别存储操作数和符号值，但是本题目不适用，我写的代码中采用了类似于求最大连续子数组的解法，设置变量一个记录当前的括号深度，一个记录最大的括号深度，从字符串头开始遍历，当遍历到字符‘('时，当前括号深度加1，当遍历到字符‘)'时，当前括号深度减1,；当当前括号深度大于最大括号深度时，将当前括号深度赋值给最大括号深度；当当前括号深度小于0时，将当前括号深度置0。算法复杂度O(n)。

代码如下：如有错误，请大家指正。

char* FindMostDeepExpression(const char* str){if(str == NULL)return NULL;int nLength = strlen(str);if(nLength < 2)return (char*)str;int maxDeepNum = 0;int tempDeepNum = 0;int startIndex = 0;int endIndex = nLength-1;bool isEnd = false; //标志是结束for(int i=0; i<nLength; ++i){if(str[i] == '('){++tempDeepNum;if(tempDeepNum > maxDeepNum ){maxDeepNum = tempDeepNum;isEnd = false;startIndex = i+1;}}else if( str[i] == ')'){if( tempDeepNum == maxDeepNum && !isEnd){endIndex = i-1;isEnd = true;}-- tempDeepNum;if(tempDeepNum < 0){tempDeepNum = 0;}}}char *result = new char[endIndex-startIndex+2];strncpy(result,str+startIndex,endIndex-startIndex+1);result[endIndex-startIndex+1] = '\0';return result;}void Test(char *testName, char* str,char* expectResult){if( testName!= NULL)printf("%s begins:\n",testName);if(expectResult != NULL)printf("expected result is %s \n",expectResult);char* result = FindMostDeepExpression(str);printf("the actual result is : %s \n",result);}void Test1(){char* str = NULL;Test("Test1",str,NULL);}void Test2(){char* str = "a+b";Test("Test2",str,"a+b");}void Test3(){char* str = "a+b+((c+d)+(e+f))";Test("Test3",str,"c+d");}void Test4(){char* str = "a+b+((c+d)+(e+f))";Test("Test3",str,"c+d");}void Test5(){char* str = "a+(b+c)";Test("Test5",str,"b+c");}int main(){Test1();Test2();Test3();Test4();system("pause");}

0 0