数据结构实验之栈四:括号匹配+(字符串中空格的输入)

    xiaoxiao2021-03-26  14

    think:1.空格的输入 2.临界情况的考虑

    数据结构实验之栈四:括号匹配 Time Limit: 1000MS Memory Limit: 65536KB

    Problem Description 给你一串字符,不超过50个字符,可能包括括号、数字、字母、标点符号、空格,你的任务是检查这一串字符中的( ) ,[ ],{ }是否匹配。

    Input 输入数据有多组,处理到文件结束。

    Output 如果匹配就输出“yes”,不匹配输出“no”

    Example Input sin(20+10) {[}]

    Example Output yes no

    Hint

    Author ma6174

    accepted

    #include <stdio.h> #include <string.h> #define MAXN 64 int top1, top2; char stacks1[MAXN], stacks2[MAXN], s[MAXN]; int main() { int i, len; while(gets(s) != NULL) { top1 = top2 = 0; len = strlen(s); for(i = 0; i < len; i++) { if(s[i] == '{' || s[i] == '[' || s[i] == '(') stacks1[top1++] = s[i]; if(s[i] == '}' || s[i] == ']' || s[i] == ')') stacks1[top1++] = s[i]; } stacks1[top1] = '\0'; stacks2[0] = stacks1[0]; for(i = 1; i < top1; ) { if((top2 >= 0) && ((stacks2[top2] == '{' && stacks1[i] == '}') || (stacks2[top2] == '[' && stacks1[i] == ']') || (stacks2[top2] == '(' && stacks1[i] == ')'))) { top2 -= 1; i++; } else { stacks2[++top2] = stacks1[i]; i++; } } if(top2 == -1 || top1 == 0) printf("yes\n"); else printf("no\n"); } return 0; } /*************************************************** User name: jk160630 Result: Accepted Take time: 0ms Take Memory: 112KB Submit time: 2017-02-01 20:21:49 ****************************************************/

    wrong answer(忽视了字符串中空格的输入)

    #include <stdio.h> #include <string.h> #define MAXN 64 int top1, top2; char stacks1[MAXN], stacks2[MAXN], s[MAXN]; int main() { int i, len; while(scanf("%s", s) != EOF) { top1 = top2 = 0; len = strlen(s); for(i = 0; i < len; i++) { if(s[i] == '{' || s[i] == '[' || s[i] == '(') stacks1[top1++] = s[i]; if(s[i] == '}' || s[i] == ']' || s[i] == ')') stacks1[top1++] = s[i]; } stacks1[top1] = '\0'; stacks2[0] = stacks1[0]; for(i = 1; i < top1; ) { if((top2 >= 0) && ((stacks2[top2] == '{' && stacks1[i] == '}') || (stacks2[top2] == '[' && stacks1[i] == ']') || (stacks2[top2] == '(' && stacks1[i] == ')'))) { top2 -= 1; i++; } else { stacks2[++top2] = stacks1[i]; i++; } } if(top2 == -1 || top1 == 0) printf("yes\n"); else printf("no\n"); } return 0; } /*************************************************** User name: jk160630 Result: Wrong Answer Take time: 0ms Take Memory: 108KB Submit time: 2017-02-01 20:18:35 ****************************************************/

    wrong answer(忽视了空格输入、未考虑临界值得带入、测试数据考虑不全面)

    #include <stdio.h> #include <string.h> #define MAXN 64 int top1, top2; char stacks1[MAXN], stacks2[MAXN], s[MAXN]; int main() { int i, len; while(scanf("%s", s) != EOF) { top1 = top2 = 0; len = strlen(s); for(i = 0; i < len; i++) { if(s[i] == '{' || s[i] == '[' || s[i] == '(') stacks1[top1++] = s[i]; if(s[i] == '}' || s[i] == ']' || s[i] == ')') stacks1[top1++] = s[i]; } stacks1[top1] = '\0'; stacks2[0] = stacks1[0]; for(i = 1; i < top1; ) { if((stacks2[top2] == '{' && stacks1[i] == '}') || (stacks2[top2] == '[' && stacks1[i] == ']') || (stacks2[top2] == '(' && stacks1[i] == ')')) { top2 -= 1; i++; } else { stacks2[++top2] = stacks1[i]; i++; } } if(top2 <= 0) printf("yes\n"); else printf("no\n"); } return 0; } /*************************************************** User name: jk160630 Result: Wrong Answer Take time: 0ms Take Memory: 112KB Submit time: 2017-02-01 10:34:22 ****************************************************/
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