leetcode兩道題

其一：Letter Combinations of a phone number

public class Solution {
public List<String> letter_Combinations(String str){
List<String> result = new ArrayList<Sting>();
if(str == null || "".equals(str))
return result;

String [] map = {" ","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
String s;
//調用遞歸方法
transfer(result , map , str , 0,s);
return result;
}

private void transfer(List<String> result,String[] map,String str,int pos, String s){
if(pos == str.length()){
result.add(s);
return;
}

char ch = str.charAt(pos);
int temp = ch-'0';
String ss  =map[temp];

if(ss == null)
transfer(result , map,str, pos+1,s);
else{
int length = ss.length();
int index;
for(index = 0 ; index < length ; index++){
char c = ss.charAt(index);
transfer(result,map,str,pos+1,s+c);
}//for
}//else
return;
}
}


//另一種風格
public class Solution{
public List<String> letterCombanations(String digits){
List<char[]> list = new ArrayList<>();
list.add(new char[] {' '});
list.add(new char[] {});
list.add(new char[] {'a','b','c'});
list.add(new char[] {'d','e','f'});
list.add(new char[] {'g','h','i'});
list.add(new char[] {'j','k','l'});
list.add(new char[] {'m','n','o'});
list.add(new char[] {'p','q','r','s'});
list.add(new char[] {'t','u','v'});
list.add(new char[] {'w','x','y','z'});

List<String> ret = new ArrayList<>();
ret.add("");
return build(ret,list,digits,0);
}

private List<String> build(List<String> ret, List<char[]> list,String digits , int pos){
if(pos == figits.length())
return ret;

int num = digits.charAt(pos)-'0';
char[] chs = list.get(num);
if(chs.length == 0)
return ret;

List<String> newRet = new ArrayList<>();
for(char c : chs){
for(String str : ret)
newRet.add(str+c);
}
return build(newRet,list,digits,pos+1);
}
}

其二：Decode ways

    我的錯誤代碼;

public int decode_misse(String str){
if(str == null || str.length() == 0)
return 0;

int num[] = new int[str.length];
num[0]=1;
int length =  str.length();
int prev = num[0]-'0';
int result = 1;
for(ini index =1; index < length ; index++){
char cur = str.charAt(index);
int temp = cur-'0';
int two  =prev*10 + temp;
if(two < 27 && two > 0)
result += num[i-1];
if(temp != 0)
result += 1;
num[index] = result;
prev = temp;
}
return num[length-1];
}
正確代碼：
public int numDecodes(String s){
if( null == s || "".equals(s))
return 0;

int[] arr = new int[s.length()];
int prev = '0';
for(int i= 0 ; i <s.length() ; i++){
int ch = s.charAt(i);
int count = 0 ;
//
int num = 10*(prev-'0')+(ch-'0');
int c = 'A'+num-1;
int prevc = 'A'+(prev-'0')-1;
if(c>='A'&&c<='Z'&&prec>='A'&&prec<='Z'){
if(i>1)
count += arr[i-2];
else if(i==1)
count += 1;
}

c = 'A'+(ch-'0')-1;
if(c >= 'A' && c <= 'Z'){
if(i > 0)
count += arr[i-1];
else count += 1;
}

if(count == 0)
return 0;

arr[i] = count;
prev = ch;
}

return arr[s.length()-1];
}