文章目录
- 216.组合总和III
- 题目描述
- 思路分析
- 代码
- 17.电话号码的字母组合
- 题目描述
- 思路分析
- 代码
216.组合总和III
题目描述
题目链接
思路分析
相对于77. 组合 (opens new window),无非就是多了一个限制,本题是要找到和为n的k个数的组合,而整个集合已经是固定的了[1,…,9]。
图中,可以看出,只有最后取到集合(1,3)和为4 符合条件。
代码
class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(int targetSum, int k, int sum, int startIndex) {if (path.size() == k) {if (sum == targetSum) {result.push_back(path);}return;}for (int i = startIndex; i <= 9; i++) {sum += i;path.push_back(i);backtracking(targetSum, k, sum, i + 1);sum -= i;path.pop_back();}}
public:vector<vector<int>> combinationSum3(int k, int n) {result.clear();path.clear();backtracking(n, k, 0, 1);return result;}
};
剪枝
class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(int targetSum, int k, int sum, int startIndex) {if (sum > targetSum) return;if (path.size() == k) {if (sum == targetSum) {result.push_back(path);}return;}for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {sum += i;path.push_back(i);backtracking(targetSum, k, sum, i + 1);sum -= i;path.pop_back();}}
public:vector<vector<int>> combinationSum3(int k, int n) {result.clear();path.clear();backtracking(n, k, 0, 1);return result;}
};
17.电话号码的字母组合
题目描述
题目链接
思路分析
问题
- 数字和字母如何映射
- 两个字母就两个for循环,三个字符我就三个for循环,以此类推
- 输入1 * #按键等等异常情况
可以使用map或者定义一个二维数组,例如:string letterMap[10],来做映射
const string letterMap[10] = {"", // 0"", // 1"abc", // 2"def", // 3"ghi", // 4"jkl", // 5"mno", // 6"pqrs", // 7"tuv", // 8"wxyz", // 9
};
例如:输入:“23”,抽象为树形结构,如图所示:
图中可以看出遍历的深度,就是输入"23"的长度,而叶子节点就是我们要收集的结果,输出[“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”]。
代码
class Solution {
private:const string letterMap[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
public:vector<string> result;string s;void backtracking(const string& digits, int index) {if (index == digits.size()) {result.push_back(s);return;}int digit = digits[index] - '0';string letters = letterMap[digit];for (int i = 0; i < letters.size(); i++) {s.push_back(letters[i]);backtracking(digits, index + 1);s.pop_back();}}vector<string> letterCombinations(string digits) {s.clear();result.clear();if (digits.size() == 0) return result;backtracking(digits, 0);;return result;}
};