0017_Letter Combinations of a Phone Number
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JAVA
方法一
要求数字对应的全部字母的组合，直接递归就可以了，不过效率不够高，大概排在中间的位置，要考虑优化。而且在写代码的时候，对substring，list的应用不熟，记不清关键字的拼写。。尴尬。。
public class Solution { public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<String>(); if(digits.length() == 0){ return result; } getString(digits,"",result); return result; } public void getString(String digits,String tempResult,List<String> result){ if(digits.length() == 0){ result.add(tempResult); return; } Character c = digits.charAt(0); switch(c){ // case '0': // getString(digits.substring(1,digits.length()),tempResult + " ",result); // break; case '2': getString(digits.substring(1,digits.length()),tempResult + "a",result); getString(digits.substring(1,digits.length()),tempResult + "b",result); getString(digits.substring(1,digits.length()),tempResult + "c",result); break; case '3': getString(digits.substring(1,digits.length()),tempResult + "d",result); getString(digits.substring(1,digits.length()),tempResult + "e",result); getString(digits.substring(1,digits.length()),tempResult + "f",result); break; case '4': getString(digits.substring(1,digits.length()),tempResult + "g",result); getString(digits.substring(1,digits.length()),tempResult + "h",result); getString(digits.substring(1,digits.length()),tempResult + "i",result); break; case '5': getString(digits.substring(1,digits.length()),tempResult + "j",result); getString(digits.substring(1,digits.length()),tempResult + "k",result); getString(digits.substring(1,digits.length()),tempResult + "l",result); break; case '6': getString(digits.substring(1,digits.length()),tempResult + "m",result); getString(digits.substring(1,digits.length()),tempResult + "n",result); getString(digits.substring(1,digits.length()),tempResult + "o",result); break; case '7': getString(digits.substring(1,digits.length()),tempResult + "p",result); getString(digits.substring(1,digits.length()),tempResult + "q",result); getString(digits.substring(1,digits.length()),tempResult + "r",result); getString(digits.substring(1,digits.length()),tempResult + "s",result); break; case '8': getString(digits.substring(1,digits.length()),tempResult + "t",result); getString(digits.substring(1,digits.length()),tempResult + "u",result); getString(digits.substring(1,digits.length()),tempResult + "v",result); break; case '9': getString(digits.substring(1,digits.length()),tempResult + "w",result); getString(digits.substring(1,digits.length()),tempResult + "x",result); getString(digits.substring(1,digits.length()),tempResult + "y",result); getString(digits.substring(1,digits.length()),tempResult + "z",result); break; } }}方法二
考虑用栈配合循环来代替递归，觉得这样可以快一些。
使用一个字符数组来模拟栈，为了减少时间，不手工初始化数组，直接使用默认初始化的值。
然而。。用时4ms，beats 47.1%。。。。印象中把递归转换成栈会节约很多时间的呀。。
public class Solution { public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<String>(); if(digits.length() == 0){ return result; } char [] temp = new char[digits.length()]; int index = 0; while(index >= 0){ if(index == digits.length()){ result.add(new String(temp)); --index; } switch (digits.charAt(index)){ case '2': switch(temp[index]){ case 0: temp[index] = 'a';++index;break; case 'a': temp[index] = 'b';++index;break; case 'b': temp[index] = 'c';++index;break; case 'c': temp[index] = 0;--index;break; } break; case '3': switch(temp[index]){ case 0: temp[index] = 'd';++index;break; case 'd': temp[index] = 'e';++index;break; case 'e': temp[index] = 'f';++index;break; case 'f': temp[index] = 0;--index;break; } break; case '4': switch(temp[index]){ case 0: temp[index] = 'g';++index;break; case 'g': temp[index] = 'h';++index;break; case 'h': temp[index] = 'i';++index;break; case 'i': temp[index] = 0;--index;break; } break; case '5': switch(temp[index]){ case 0: temp[index] = 'j';++index;break; case 'j': temp[index] = 'k';++index;break; case 'k': temp[index] = 'l';++index;break; case 'l': temp[index] = 0;--index;break; } break; case '6': switch(temp[index]){ case 0: temp[index] = 'm';++index;break; case 'm': temp[index] = 'n';++index;break; case 'n': temp[index] = 'o';++index;break; case 'o': temp[index] = 0;--index;break; } break; case '7': switch(temp[index]){ case 0: temp[index] = 'p';++index;break; case 'p': temp[index] = 'q';++index;break; case 'q': temp[index] = 'r';++index;break; case 'r': temp[index] = 's';++index;break; case 's': temp[index] = 0;--index;break; } break; case '8': switch(temp[index]){ case 0: temp[index] = 't';++index;break; case 't': temp[index] = 'u';++index;break; case 'u': temp[index] = 'v';++index;break; case 'v': temp[index] = 0;--index;break; } break; case '9': switch(temp[index]){ case 0: temp[index] = 'w';++index;break; case 'w': temp[index] = 'x';++index;break; case 'x': temp[index] = 'y';++index;break; case 'y': temp[index] = 'z';++index;break; case 'z': temp[index] = 0;--index;break; } break; } } return result; }}阅读全文
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