LEET code -- Bulls and Cows

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.
For example:

Secret number:  "1807"
Friend's guess: "7810"

Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.)

Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows. In the above example, your function should return "1A3B".
Please note that both secret number and friend's guess may contain duplicate digits, for example:

Secret number:  "1123"
Friend's guess: "0111"

In this case, the 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return "1A1B".


You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.






要注意char的用法

char * f = "abcdef";
strcpy(f, "abcdef");
printf("%s",f);
//上面會掛  因為當char *指向一串"" 的字串  會把它處理為 唯讀

The first example is a char * to a character literal (a literal is "something"). Character literals are read-only, and attempting to write to them can result in crashes. Your first pointer should really be const char *f = "abcdef";, which strcpy won't take.

char s[] = "ddd";
strcpy(&s[0], "eee");
printf("%s", s);

//這個會放在stack  所以是可改的

http://stackoverflow.com/questions/5645949/program-aborts-when-using-strcpy-on-a-char-pointer-works-fine-on-char-array




要注意 strcpy一定要是在stack的獲釋malloc 這些有實體記憶體位置的
不能直接char *p;  一個空指標過去

另外 在第26是傳&p[0]回去，這樣不行是因為 p是stack的區域變數  回傳回去記憶體就被釋放了  必須要一個在heap的記憶體ANS來負責回傳

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char* getHint(char* secret, char* guess) { int size=strlen(secret); int A=0,B=0; char p[size]; //strcpy(p,guess); // can't use pointer in strcpy *p strcpy(&p[0],guess); for(int i=0;i<size;i++)if(secret[i]==guess[i]){ A++; p[i]='A'; } for(int i=0;i<size;i++){ for(int j=0;j<size;j++){ if(secret[i]==p[j] &&p[i]!='A' ){//p[i]!='A' is to avoid, the (1122,1222), need to avoid the A B++; p[j]='b'; break; } } } char *ans=malloc(sizeof(char)*size); sprintf(ans,"%dA%dB\0",A,B); //sprintf(p,"%dA%dB\0",A,B); printf("%s",p); //return &p[0]; return ans;//can't use &p[0], because the p is in the stack }

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更漂亮一點  不需要額外跟guess 依樣長的字串
方法是用一個HASH array (第4行)
裡面記錄誰看過這個數字了  
+1 代表secret看過
-1  代表guess看過
所以當secret檢查發現<0 guess="" nbsp="" p="" secret="">當guess檢查的時候發現>0  代表之前secret有放進來 ，可以算進B

要注意的是第5行的memset 只能用在初始0 或是 -1  因為他是把所以的記憶體都寫成一樣的內容

還有要注意11行要減去'0'  才可以轉換成0~9的範圍

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char* getHint(char* secret, char* guess) { int A=0,B=0; int p[10]; memset(&p[0],0,sizeof(int)*10);//beware the memset can only set to 0 or -1 for(;*secret;secret++,guess++){ if(*secret==*guess){ A++; }else{ if(p[(int)*secret -'0'] <0 )B++; //if is <0, there has a guess put a value in here before p[(int)*secret-'0']+=1;//use 1 to indicate the secret has a vaule in here if(p[(int)*guess -'0'] >0)B++;//if >0, there is a secret put a value in here p[(int)*guess-'0']-=1; //use -1 to indicate the guess has a value in here printf("%d %d %d\n",p[(int)*secret-'0'],p[(int)*guess-'0'],B); } } char *ans=malloc(sizeof(char)*4); sprintf(ans,"%dA%dB\0",A,B); //sprintf(p,"%dA%dB\0",A,B); printf("%s",p); //return &p[0]; return ans;//can't use &p[0], because the p is in the stack }