216.组合总和III
题目链接：216. 组合总和 III - 力扣（LeetCode）
思路：
class Solution {
List<List<Integer>> result = new ArrayList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> combinationSum3(int k, int n) {
combinationSum4(k,n,1,0);
return result;
}
public void combinationSum4(int k, int n, int start, int sum){
if(sum > n){
return;
}
if(path.size() == k){
if(sum == n){
result.add(new ArrayList<>(path));
return;
}
}
//i的最大值和path集合个数有关，第一个数最大值为7，第二个最大值为8，第三个最大值9
//到这一步时，path.size()最大值为2，
for(int i = start; i <= 9 - (k - path.size()) + 1; i++ ){
path.add(i);
//sum += i;
combinationSum4(k,n,i+1,sum+i);
path.removeLast();
//sum -=i;
}
return;
}
}17.电话号码的字母组合
思路链接：17. 电话号码的字母组合 - 力扣（LeetCode）
思路：难点在于字符串处理
class Solution {
List<String> list = new ArrayList<>();
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;
}
StringBuilder temp = new StringBuilder();
public void backTracking(String digits,String[] numString, int num){
if(num == digits.length()){
list.add(temp.toString());
return;
}
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.deleteCharAt(temp.length() -1);
}
}
}
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