h*d
2 楼
没有人知道吗？
j*t
3 楼
这是trie吗? 应该是一个recursive search吧.
s*y
4 楼
static final int phoneLength=7;
char [][] keypad={{'0'},{'1'},{'a','b','c'},{'d','e','f'},
{'g','h','i'},{'j','k','l'},{'m','n','o'},
{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
void printWords(int [] phoneNumber){
char result []= new char [phoneLength];
doPrintWords(phoneNumber,result,0);
}
void doPrintWords(int []phoneNumber, char []result, int curIndex){
if(curIndex==phoneLength){
System.out.println(result);
return;
}
if(phoneNumber[curIndex]==0){
result[curIndex]='0';
doPrintWords(phoneNumber,result, curIndex+1);
}
else if(phoneNumber[curIndex]==1){
result[curIndex]='1';
doPrintWords(phoneNumber,result, curIndex+1);
}
else{
int numOfchar=keypad[phoneNumber[curIndex]].length;
for(int i=0;i
doPrintWords(phoneNumber,result, curIndex+1);
}
}
char [][] keypad={{'0'},{'1'},{'a','b','c'},{'d','e','f'},
{'g','h','i'},{'j','k','l'},{'m','n','o'},
{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
void printWords(int [] phoneNumber){
char result []= new char [phoneLength];
doPrintWords(phoneNumber,result,0);
}
void doPrintWords(int []phoneNumber, char []result, int curIndex){
if(curIndex==phoneLength){
System.out.println(result);
return;
}
if(phoneNumber[curIndex]==0){
result[curIndex]='0';
doPrintWords(phoneNumber,result, curIndex+1);
}
else if(phoneNumber[curIndex]==1){
result[curIndex]='1';
doPrintWords(phoneNumber,result, curIndex+1);
}
else{
int numOfchar=keypad[phoneNumber[curIndex]].length;
for(int i=0;i
doPrintWords(phoneNumber,result, curIndex+1);
}
}
h*d
5 楼
这个是print电话对应的字母吧？
【在 s********y 的大作中提到】
: static final int phoneLength=7;
: char [][] keypad={{'0'},{'1'},{'a','b','c'},{'d','e','f'},
: {'g','h','i'},{'j','k','l'},{'m','n','o'},
: {'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
: void printWords(int [] phoneNumber){
: char result []= new char [phoneLength];
: doPrintWords(phoneNumber,result,0);
: }
: void doPrintWords(int []phoneNumber, char []result, int curIndex){
: if(curIndex==phoneLength){
【在 s********y 的大作中提到】
: static final int phoneLength=7;
: char [][] keypad={{'0'},{'1'},{'a','b','c'},{'d','e','f'},
: {'g','h','i'},{'j','k','l'},{'m','n','o'},
: {'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
: void printWords(int [] phoneNumber){
: char result []= new char [phoneLength];
: doPrintWords(phoneNumber,result,0);
: }
: void doPrintWords(int []phoneNumber, char []result, int curIndex){
: if(curIndex==phoneLength){
l*a
6 楼
每个节点保存所有相关人名list.
struct node
{
int number;
struct node* sub[10];
struct list * head;
struct list * tail;
}
struct list
{
char * name;
struct list * next;
}
剩下的就不用了吧？
【在 h**********d 的大作中提到】
: “手机上给了电话号码求人名索引”
: 谁有这题做法的code，能不能分享一下，非常感谢！
struct node
{
int number;
struct node* sub[10];
struct list * head;
struct list * tail;
}
struct list
{
char * name;
struct list * next;
}
剩下的就不用了吧？
【在 h**********d 的大作中提到】
: “手机上给了电话号码求人名索引”
: 谁有这题做法的code，能不能分享一下，非常感谢！
h*d
7 楼
thanx!
【在 l*****a 的大作中提到】
: 每个节点保存所有相关人名list.
: struct node
: {
: int number;
: struct node* sub[10];
: struct list * head;
: struct list * tail;
: }
: struct list
: {
【在 l*****a 的大作中提到】
: 每个节点保存所有相关人名list.
: struct node
: {
: int number;
: struct node* sub[10];
: struct list * head;
: struct list * tail;
: }
: struct list
: {
c*2
8 楼
This is a lookup problem.
To use prefix tree for this: build the tree with 0-9, at the end node, store
person's name.
(This is similar to routing table lookup.)
Can also be done with hash mapping.
To use prefix tree for this: build the tree with 0-9, at the end node, store
person's name.
(This is similar to routing table lookup.)
Can also be done with hash mapping.
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