Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
- Only numbers
1through9are used. - Each number is used at most once.
Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.
Example 2:
Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.
Example 3:
Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.
Constraints:
2 <= k <= 91 <= n <= 60
回溯法，不难写。剪枝操作绕了一会。结果是用 9 - 当前元素 + 1 > k - path.size()，即可用的数字 > 需要的数字来写，且判断发生在将当前循环元素加入递归之前，所以带个 +1
class Solution {
public void backtracking(LinkedList<Integer> path, int k, int n, int sum, int start, List<List<Integer>> res) {
if(path.size() == k) {
if(sum == n) {
res.add(new ArrayList<Integer>(path));
}
return;
}
for(int i=start; i<=9 && (9 - i + 1) >= (k - path.size()); i++) {
path.add(i);
backtracking(path, k, n, sum + i, i+1, res);
path.removeLast();
}
}
public List<List<Integer>> combinationSum3(int k, int n) {
LinkedList<Integer> path = new LinkedList<Integer>();
List<List<Integer>> res = new LinkedList<List<Integer>>();
backtracking(path, k, n, 0, 1, res);
return res;
}
}
17. Letter Combinations of a Phone Number
Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example 1:
Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
Example 2:
Input: digits = ""
Output: []
Example 3:
Input: digits = "2"
Output: ["a","b","c"]
Constraints:
0 <= digits.length <= 4digits[i]is a digit in the range['2', '9'].
这题感觉写麻烦了一些，如果定义一个字符串数组，把数值map到可选字符串会简单一些。 但用switch case是比较容易处理异常字符。代码本身没考虑异常字符，考虑的话，要把default的处理改动下，移到上面的case里。用default逻辑来处理异常字符。
class Solution {
public void backtracking(StringBuilder sb, int pos, char[] digits, List<String> res) {
if(sb.length() == digits.length) {
if(sb.length() > 0) {
res.add(new String(sb.toString()));
}
return;
}
int cur = digits[pos] - '0'; //get digit number
if(cur == 1) {
//skip current
backtracking(sb, pos+1, digits, res);
}
else if (cur>=2 && cur<=9) {
//23456789
int num =0;
int start = 0;
switch(cur) {
case 7:
num = 4;
start = 15;
break;
case 8:
num = 3;
start = 19;
break;
case 9:
num = 4;
start = 22;
break;
default:
num = 3;
start = (cur-2) * 3;
}
for(int i=0; i<num; i++) {
char cc = (char)('a' + start + i);
sb.append(cc);
backtracking(sb, pos+1, digits, res);
sb.deleteCharAt(sb.length() - 1);
}
}
else {
return;
}
}
public List<String> letterCombinations(String digits) {
List<String> res = new ArrayList<String>();
char[] ad = digits.toCharArray();
StringBuilder sb = new StringBuilder();
System.out.println(ad.length);
backtracking(sb, 0, ad, res);
return res;
}
}
