原题链接：

17. 电话号码的字母组合

https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/

完成情况：

解题思路：

没啥思路，就是常规回溯，多多模仿。理解即可。

参考代码：

class Solution {List<String> result = new ArrayList<String>();//构造一个String类去存储每一个子集,考虑到String使用有限，我们采用sbStringBuilder subSets = new StringBuilder();//对于本题，还要绘制一个phoneButtonMapMap<Character,String> phoneButtonMap = new HashMap<Character,String>(){{put('2',"abc");put('3',"def");put('4',"ghi");put('5',"jkl");put('6',"mno");put('7',"pqrs");put('8',"tuv");put('9',"wxyz");}};/**** @param digits* @return*/public List<String> letterCombinations(String digits) {//判断非空，然后进入递归if (digits == null||digits.length() == 0   ){return result;}bactTrack(digits,0,subSets);return result;}/*** * @param digits* @param startIndex* @param subSets*/private void bactTrack(String digits, int startIndex, StringBuilder subSets) {//先给出终止条件，在写进去回溯情况if (startIndex == digits.length()){result.add(subSets.toString());return;}//获取字符，然后对字符里面的每一个数都去尝试一下char dight = digits.charAt(startIndex);StringBuilder stringBuilder = new StringBuilder(phoneButtonMap.get(dight));for (int i=0;i<stringBuilder.length();i++){//获取Map对应的每一个字符组成一种情况,先加入到subSets列表subSets.append(stringBuilder.charAt(i));bactTrack(digits,startIndex+1,subSets);subSets.deleteCharAt(subSets.length()-1);//最后通过递归，再删除尾部节点}}
}

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