代码随想录算法训练营第二十五天| 216 组合总合 ||| 17 电话号码的字母组合

216 组合总和 |||

暴力

class Solution {List<List<Integer>>res = new ArrayList<>();List<Integer>newList = new ArrayList<>();public List<List<Integer>> combinationSum3(int k, int n) {soluHelper(1,k,n,0);return res;}private void soluHelper(int num,int k,int n,int sum){if(sum > n)return;//减枝操作if(newList.size() == k){if(sum == n)res.add(new ArrayList(newList));return;}for(int i = num;i <= 9;i++){newList.add(i);sum += i;soluHelper(i + 1,k,n,sum);newList.remove(newList.size() - 1);sum -= i;}}
}

时间复杂度O(N × $2^{N}$ )N为集合的大小，本题中为9，一个有 $2^{N}$ 个状态(选择或者不选这个数字)

空间复杂度O(N) newList的空间大小

减枝优化

class Solution {List<List<Integer>>res = new ArrayList<>();List<Integer>newList = new ArrayList<>();public List<List<Integer>> combinationSum3(int k, int n) {soluHelper(1,k,n,0);return res;}private void soluHelper(int num,int k,int n,int sum){if(sum > n)return;//减枝操作if(newList.size() == k){if(sum == n)res.add(new ArrayList(newList));return;}for(int i = num;i <= 9 - (k - newList.size()) + 1;i++){//减枝操作newList.add(i);sum += i;soluHelper(i + 1,k,n,sum);newList.remove(newList.size() - 1);sum -= i;}}
}

时间复杂度O(N × $2^{N}$ )N为集合的大小，本题中为9，一个有 $2^{N}$ 个状态(选择或者不选这个数字)

空间复杂度O(N) newList的空间大小

17 电话号码的字母组合

class Solution {List<String>res = new ArrayList<>();StringBuffer sb = new StringBuffer();Map<Integer,String>mp = new HashMap<>();String[] strs = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};public List<String> letterCombinations(String digits) {if(digits.length() == 0 || digits == null)return res;SoluHelper(digits,0);return res;}private void SoluHelper(String digits,int cnt){if(cnt == digits.length()){res.add(sb.toString());return;}String str = strs[digits.charAt(cnt) - '0'];for(int i = 0;i < str.length();i++){sb.append(str.charAt(i));SoluHelper(digits,cnt + 1);sb.deleteCharAt(sb.length() - 1);}}
}

时间复杂度O( $3^{N}$ × $4^{M}$ )N是对应三个字母的数字个数,M是对应四个字母的数字个数