leetcode439三元表达式解析器

秒速五厘米 2022-10-09 12:21 159阅读 0赞

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思路：**使用栈，并从最右开始**

1.  当遇到?时，将栈顶的两个元素取出，与?的左一位进行T，F比对
2.  如果不是？或 :，那么元素入栈
3.  注意结果不一定是数值，还有可能是字母，所以入栈的不会全是数字，字母T，F也会入栈

代码

char * parseTernary(char * a){ 
        char *stack=(char *)malloc(sizeof(char)*10000);
        int len = strlen(a),i,s=0;
        for(i=len-1;i>=0;i--){ 
            if(a[i]=='?'){ 
                char t1 = stack[--s];
                char t2 = stack[--s];
                if(a[--i]=='T')//--i这个地方与for循环的i--跳过了T
                    stack[s++] = t1;
                else stack[s++] = t2; 
            }
            else if(a[i]!=':')//数值入栈
                stack[s++] = a[i];
        }
        stack[1]='\0';
    
        return stack;
    }

char * parseTernary(char * a){ 
        int len = strlen(a);
        int i =len -1;
        char *stack = (char*)malloc(sizeof(char)*10000);
        int s = 0;
        while (i>=0)
        { 
            if(a[i]!=':'&&a[i]!='?'){ 
                stack[s++]=a[i];
            }
            else if(a[i]=='?'){ 
                char t1 = stack[--s];
                char t2 = stack[--s];
                if(a[--i]=='T')
                    stack[s++] = t1;
                else 
                    stack[s++] = t2;
            }
            i--;
        }
    
        stack[1]='\0';
    
        return stack;
        
    }

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