代码随想录Day28

17.电话号码的字母组合

题目：17. 电话号码的字母组合 - 力扣（LeetCode）

思路：有点难，每个数字对应几个字母，首先要建立这种映射，之后才能根据传进去的digits找到要用哪些字母集进行组合，它也要用回溯，应该就是每个数字对应的字母集中取一个，然后找出所有这样的组合

尝试

class Solution {List<String> result = new ArrayList<>();StringBuilder s = new StringBuilder();Map<Integer,String> map = new HashMap<>();public List<String> letterCombinations(String digits) {map.put(2,"abc"); map.put(3,"def");map.put(4,"ghi");map.put(5,"jkl");map.put(6,"mno");map.put(7,"pqrs");map.put(8,"tuv");map.put(9,"wxyz");String[] digitsSplit = digits.split(""); // 将字符串分割成单个字符int[] digitsArray = new int[digitsSplit.length];for (int i = 0; i < digitsSplit.length; i++) {digitsArray[i] = Integer.parseInt(digitsSplit[i]); // 将每个字符转换为整数}return result;}private void combine(int[] array,int startIndex){if(s.length() == array.size()){result.add(new String(s));return;}for(int i = startIndex; i <= array.size();i++ ){// 这里应该是要根据array里面的数字，取出map里面的字母集，后面不会写了s.append(i);combine(array,startIndex);s.removeLast();}}
}

答案

class Solution {//设置全局列表存储最后的结果List<String> list = new ArrayList<>();public List<String> letterCombinations(String digits) {if (digits == null || digits.length() == 0) {return list;}//初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};//迭代处理backTracking(digits, numString, 0);return list;}//每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuilderStringBuilder temp = new StringBuilder();//比如digits如果为"23",num 为0，则str表示2对应的 abcpublic void backTracking(String digits, String[] numString, int num) {//遍历全部一次记录一次得到的字符串if (num == digits.length()) {list.add(temp.toString());return;}//str 表示当前num对应的字符串String str = numString[digits.charAt(num) - '0'];for (int i = 0; i < str.length(); i++) {temp.append(str.charAt(i));//cbackTracking(digits, numString, num + 1);//剔除末尾的继续尝试temp.deleteCharAt(temp.length() - 1);}}
}

默写

class Solution {List<String> list = new ArrayList<>();StringBuilder temp = new StringBuilder();public List<String> letterCombinations(String digits) {if(digits == null || digits.length() == 0) return list;String[] numString = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};backTracking(digits,numString,0); return list;}public void backTracking(String digits,String[] numString,int num){// if(temp.size() == digits.size()){//     list.add(temp.toString());//     return;// }if(num == digits.length()){list.add(temp.toString());return;}// String[] str = numString[digits.charAt(num) - '0'];String str = numString[digits.charAt(num) - '0'];for(int i = 0;i < str.length();i++){temp.append(str.charAt(i));backTracking(digits,numString,num +1);// temp.removeCharAt(temp.length()-1);temp.deleteCharAt(temp.length()-1);}}
}