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<int>> combine(int n, int k) {
    backTracking(n, k, 1);
    return result;
}
vector<int> temp;
vector<vector<int>> 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<int>> 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<string> letterCombinations(string digits) {
    string lToNMap[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    if(digits == "")
    {
        return result;
    }
    backTracking(digits, 0, lToNMap);
    return result;

}

};`