(C++回溯01) 组合

77、组合

回溯题目三步走

1. 确定参数

2. 确定终止条件

3. for 循环横向遍历，递归纵向遍历

class Solution {
public:vector<vector<int>> result;vector<int> path;void backtracking(int n, int k, int startIndex) {if(path.size() == k) {result.push_back(path);return;}for(int i = startIndex; i <= n; i++) {path.push_back(i);backtracking(n, k, i + 1);path.pop_back();}}vector<vector<int>> combine(int n, int k) {backtracking(n, k, 1);return result;}
};

剪枝操作：

公式：n - (k - path.size()) + 1

k - path.size() 意思是还可选取多少元素

n - (k - path.size()) 意思是去掉可选取元素个数后的位置

n - (k - path.size()) + 1 为当前可以组合的最后一个位置

剪枝后代码：

class Solution {
public:vector<vector<int>> result;vector<int> path;void backtracking(int n, int k, int startIndex) {if(path.size() == k) {result.push_back(path);return;}for(int i = startIndex; i <= n; i++) {int flag = n - (k - path.size()) + 1;if(i > flag) {return;}path.push_back(i);backtracking(n, k, i + 1);path.pop_back();}}vector<vector<int>> combine(int n, int k) {backtracking(n, k, 1);return result;}
};

216、组合总和 III

class Solution {
public:vector<vector<int>> result;vector<int> path;int count = 0;void backtracking(int k, int n, int startIndex) {if(path.size() == k) {if(count == n)result.push_back(path);return;}for(int i = startIndex; i < 10; i++) {path.push_back(i);count += i;backtracking(k, n, i + 1);path.pop_back();count -= i;}}vector<vector<int>> combinationSum3(int k, int n) {backtracking(k, n, 1);return result;}
};

剪枝操作：

class Solution {
public:vector<vector<int>> result;vector<int> path;int count = 0;void backtracking(int k, int n, int startIndex) {if(path.size() == k) {if(count == n)result.push_back(path);return;}for(int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {count += i;if(count > n) {count -= i;return;}path.push_back(i);backtracking(k, n, i + 1);path.pop_back();count -= i;}}vector<vector<int>> combinationSum3(int k, int n) {backtracking(k, n, 1);return result;}
};

17、电话号码的字母组合

class Solution {
private:string letterMap[10] = {"","","abc", // 2"def", // 3"ghi", // 4"jkl", // 5"mno", // 6"pqrs", // 7"tuv", // 8"wxyz", // 9};
public:vector<string> result;string path;void backtacking(string& digits, int index) {if(index == digits.size()) {result.push_back(path);return;}string letters = letterMap[digits[index] - '0'];for(int i = 0; i < letters.size(); i++) {path.push_back(letters[i]);backtacking(digits, index + 1);path.pop_back();}}vector<string> letterCombinations(string digits) {if(digits.size() == 0) {return result;}backtacking(digits, 0);return result;}
};