回溯第一天

回溯算法理论基础
题目分类：组合 分割 子集 排列 棋盘 其他
模板
`

void backtracking(参数) {
if (终止条件) {
存放结果;
return;
}
for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
处理节点;
backtracking(路径，选择列表); // 递归
回溯，撤销处理结果
}
}`

组合
`

vector<vector> combine(int n, int k) {
backTracking(n, k, 1);
return result;
}
vector temp;
vector<vector> result;
void backTracking(int n, int k, int startIndex)
{
// if(n - temp.size() )
if(temp.size() == k)
{
result.push_back(temp);
return;
}
for(int i = startIndex; i <= n - (k - temp.size()) + 1; i++)
{
temp.push_back(i);
backTracking(n, k, i + 1);
temp.pop_back();
}
}

`

** 组合总和 III**

`
class Solution {
public:
vector<vector> result;
vector temp;
void backTracking(int k, int n, int startIndex, int sum)
{
if(sum > n )
return;
if(sum == n && temp.size() == k)
{
result.push_back(temp);
return;
}
for(int i = startIndex; i <= 9 - (k - temp.size()) + 1; i++ )
{
temp.push_back(i);
sum += i;
backTracking(k, n, i + 1, sum);
sum -= i;
temp.pop_back();
}

}
vector<vector> combinationSum3(int k, int n) {
int sum = 0;
backTracking(k, n, 1, sum);
return result;
}
};
电话号码的字母组合class Solution {
public:
vector result;
string temp = "";
void backTracking(string digits, int startIndex, string lToNMap[])
{
if(temp.size() == digits.size())
{
result.push_back(temp);
return;
}

int index = (int)(digits[startIndex] - '0');
for(int i = 0; i < lToNMap[index].size(); i++)
{
temp += lToNMap[index][i];
backTracking(digits, startIndex + 1, lToNMap);
temp.pop_back();
}
}
vector letterCombinations(string digits) {
string lToNMap[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
if(digits == "")
{
return result;
}
backTracking(digits, 0, lToNMap);
return result;

}
};`