216. Combination Sum III
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
在[1,2,3,4,5,6,7,8,9]这个集合中找到和为n的k个数的组合。
k相当于了树的深度，9（因为整个集合就是9个数）就是树的宽度。
需要一维数组path来存放符合条件的结果，二维数组result来存放结果集。
还需要如下参数：
- targetSum（int）目标和，也就是题目中的n。
- k（int）就是题目中要求k个数的集合。
- sum（int）为已经收集的元素的总和，也就是path里元素的总和。
- startIndex（int）为下一层for循环搜索的起始位置。
k其实就已经限制树的深度，因为就取k个元素，树再往下深了没有意义。
如果path.size() 和 k相等了，就终止
如果此时path里收集到的元素和（sum） 和targetSum（就是题目描述的n）相同了，就用result收集当前的结果。
处理过程 和 回溯过程是一一对应的，处理有加，回溯就要有减！
已选元素总和如果已经大于n（图中数值为4）了，那么往后遍历就没有意义了，直接剪掉。
class Solution {
LinkedList<Integer> path = new LinkedList<>();
List<List<Integer>> ans = new ArrayList<>();
public List<List<Integer>> combinationSum3(int k, int n) {
build(k, n, 1, 0);
return ans;
}
private void build(int k, int n, int startIndex, int sum) {
if (sum > n) return;
if (path.size() > k) return;
if (sum == n && path.size() == k) {
ans.add(new ArrayList<>(path));
return;
}
for(int i = startIndex; i <= 9; i++) {
path.add(i);
sum += i;
build(k, n, i + 1, sum);
sum -= i;
path.removeLast();
}
}
}
For Future References
题目链接：https://leetcode.com/problems/combination-sum-iii/
文章讲解：https://programmercarl.com/0216.组合总和III.html
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'].
首先需要一个字符串s来收集叶子节点的结果，然后用一个字符串数组result保存起来
index是记录遍历第几个数字了，就是用来遍历digits的（题目中给出数字字符串），同时index也表示树的深度。
终止条件就是如果index 等于 输入的数字个数（digits.size）了（本来index就是用来遍历digits的）。
然后收集结果，结束本层递归。
首先要取index指向的数字，并找到对应的字符集（手机键盘的字符集）。
然后for循环来处理这个字符集
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 StringBuild
StringBuilder temp = new StringBuilder();
//比如digits如果为"23",num 为0，则str表示2对应的 abc
public 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));
//c
backTracking(digits, numString, num + 1);
//剔除末尾的继续尝试
temp.deleteCharAt(temp.length() - 1);
}
}
}
For Future References
题目链接：https://leetcode.com/problems/letter-combinations-of-a-phone-number/
文章讲解：https://programmercarl.com/0017.电话号码的字母组合.html
标签：216,25,return,int,sum,随想录,digits,combinations,path From： https://www.cnblogs.com/bluesociety/p/16793450.html