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A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:
- S is empty;
- S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;
- S has the form "VW" where V and W are properly nested strings.
For example, the string "{[()()]}" is properly nested but "([)()]" is not.
Write a function:
int solution(char *S);
that, given a string S consisting of N characters, returns 1 if S is properly nested and 0 otherwise.
For example, given S = "{[()()]}", the function should return 1 and given S = "([)()]", the function should return 0, as explained above.
Assume that:
- N is an integer within the range [0..200,000];
- string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
Copyright 2009–2015 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
a.每次peek后，需要检查时候为NULL,如果缺少这一步检查而直接进行 myNode->x 操作，会产生段错误。
b.注意str的对称性问题，并不是每一次循环正常结束都是正确的，有可能存在右括号少的情况，需要判断如果stack不为空，则返回false；
c.要注意temp必须每一次都++,否则循环不能正常进行。
1 // you can write to stdout for debugging purposes, e.g. 2 // printf("this is a debug message\n"); 3 #include <stdlib.h> 4 5 typedef struct node{ 6 char x; 7 struct node* next; 8 }tnode; 9 10 typedef struct stack{ 11 tnode *top; 12 tnode *bottom; 13 }tStack; 14 15 void push(tnode *N,tStack *S) 16 { 17 if(S->top == NULL) 18 { 19 S->top = N; 20 S->bottom = N; 21 } 22 else 23 { 24 N->next = S->top; 25 S->top = N; 26 } 27 } 28 29 tnode *peek(tStack *S) 30 { 31 if(S->top == NULL) 32 { 33 return NULL; 34 } 35 else 36 { 37 return S->top; 38 } 39 } 40 41 tnode *pop(tStack *S) 42 { 43 if(S->top == NULL) 44 { 45 return NULL; 46 } 47 if(S->top == S->bottom) 48 { 49 tnode *temp = S->top; 50 S->top = NULL; 51 S->bottom = NULL; 52 return temp; 53 } 54 else 55 { 56 tnode *temp = S->top; 57 S->top = S->top->next; 58 return temp; 59 } 60 } 61 62 int solution(char *S) { 63 // write your code in C99 64 65 tStack *myStack = malloc(sizeof(tStack)); 66 myStack->top = NULL; 67 myStack->bottom = NULL; 68 69 // int len = strlen(S); 70 char *temp = S; 71 tnode *myNode = NULL; 72 // int i; 73 while(*temp) 74 { 75 // printf("%c\n",*temp); 76 if(*temp == ‘(‘ || *temp == ‘[‘ ||*temp == ‘{‘) 77 { 78 myNode = malloc(sizeof(tnode)); 79 myNode->x = *temp; 80 push(myNode,myStack); 81 } 82 else 83 { 84 myNode = peek(myStack); 85 if(myNode == NULL) 86 { 87 return 0; 88 } 89 if(*temp == ‘)‘) 90 { 91 if(myNode->x == ‘(‘) 92 { 93 myNode = pop(myStack); 94 } 95 else 96 { 97 return 0; 98 } 99 } 100 if(*temp == ‘]‘) 101 { 102 if(myNode->x == ‘[‘) 103 { 104 myNode = pop(myStack); 105 } 106 else 107 { 108 return 0; 109 } 110 } 111 if(*temp == ‘}‘) 112 { 113 if(myNode->x == ‘{‘) 114 { 115 myNode = pop(myStack); 116 } 117 else 118 { 119 return 0; 120 } 121 } 122 } 123 124 temp++; 125 } 126 myNode = peek(myStack); 127 if(myNode == NULL) 128 { 129 return 1; 130 } 131 else 132 { 133 return 0; 134 } 135 136 }
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原文地址：http://www.cnblogs.com/sjtubear/p/4233112.html
