标签：
https://leetcode.com/problems/palindrome-partitioning/
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s =
"aab",
Return[ ["aa","b"], ["a","a","b"] ]
/**
* author ： Jianxin Zhou
* email：zhoujx0219@163.com
*
* 该题dfs函数原型如下：
* void partitionHelper(const string &s, vector<vector<string>> &result, vector<string> &path, int pos)
*
* 以aaba举例。
* 1. 首先a为回文，然后对aba进行dfs
* 2. 之后回溯到a时，以aa为回文，然后对ba做dfs
* 3. 回溯到aa，试图以aab为回文，失败；试图以aaba为回文失败；结束。
*
* 注意：如果能顺利的找到一组回文，那么pos最终会等于s.size(),此时可以push到result。
* 如果找不到，例如之前的aaba不是回文，那么就会直接退出循环，没有机会执行下一步递归，也就没有pos等于s.size了。
*
* 实际上，此类题与真正的dfs的差别在于，dfs在回溯时，不会进行剪枝操作。而此类题，由于需要求出所有方案，所以需要剪枝。
*
*/
class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> result;
vector<string> path;
partitionHelper(s, result, path, 0);
return result;
}
private:
void partitionHelper(const string &s, vector<vector<string>> &result, vector<string> &path, int pos) {
// base case
if (pos == s.size()) {
result.push_back(path);
return;
}
for (int i = pos; i < s.size(); i++) {
if (isPalindrome(s, pos, i)) {
path.push_back(s.substr(pos, i - pos + 1));
partitionHelper(s, result, path, i + 1);
path.pop_back();
}
}
}
bool isPalindrome(const string &s, int start, int end) {
while (start < end) {
if (s[start] == s[end]) {
start++;
end--;
} else {
break;
}
}
return start >= end;
}
};
https://leetcode.com/problems/permutations/
Given a collection of numbers, return all possible permutations.
For example,
[1,2,3]have the following permutations:[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2], and[3,2,1].
具体可参加我之前写的文章：[LintCode] Permutations
/**
* 思路：dfs。
*
* 以123举例，
* 1. 首先以1作为head，然后对23做dfs
* 2. 回溯到1， 以2作为head，对13做dfs
* 3. 最后回溯到2，以3作为head，对12做dfs
*
* 注意：例如以2为head，对其余元素做dfs时，那么2不能再取，因此在进行下一轮dfs时，需要标记2为以访问过
*
*/
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> result;
vector<int> path;
bool visited[nums.size()];
for(int i = 0; i < nums.size(); i++) {
visited[i] = false;
}
sort(nums.begin(), nums.end());
dfs(nums, result, path, visited);
return result;
}
private:
void dfs(const vector<int> &nums, vector<vector<int>> &result, vector<int> &path, bool visited[]) {
// base case
if (path.size() == nums.size()) {
result.push_back(path);
return;
}
for (int i = 0; i < nums.size(); i++) {
if (visited[i] == false) {
path.push_back(nums[i]);
visited[i] = true;
dfs(nums, result, path, visited);
path.pop_back();
visited[i] = false;
}
}
}
};
https://leetcode.com/problems/permutations-ii/
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2]have the following unique permutations:[1,1,2],[1,2,1], and[2,1,1].
要点在于保证相同的数不在同一位置出现两次以上，可以参见我写的这篇文章：[LintCode] Permutations II
class Solution {
public:
/**
* @param nums: A list of integers.
* @return: A list of unique permutations.
*/
vector<vector<int> > permuteUnique(vector<int> &nums) {
// write your code here
vector<vector<int>> paths;
if (nums.empty()) {
return paths;
}
sort(nums.begin(), nums.end());
bool *visited = new bool[nums.size()]();
vector<int> path;
permuteUniqueHelper(nums, visited, path, paths);
return paths;
}
private:
void permuteUniqueHelper(const vector<int> &nums,
bool visited[],
vector<int> &path,
vector<vector<int>> &paths) {
if (path.size() == nums.size()) {
paths.push_back(path);
return;
}
for (int ix = 0; ix < nums.size(); ix++) {
if (visited[ix] == true || ix > 0 && nums[ix - 1] == nums[ix] && visited[ix - 1] == false) {
continue;
}
visited[ix] = true;
path.push_back(nums[ix]);
permuteUniqueHelper(nums, visited, path, paths);
visited[ix] = false;
path.pop_back();
}
}
};
https://leetcode.com/problems/subsets/
Given a set of distinct integers, nums, return all possible subsets.
Note:
- Elements in a subset must be in non-descending order.
- The solution set must not contain duplicate subsets.
For example,
If nums =[1,2,3], a solution is:[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
/**
* 思路：找方案，一般都是使用搜索。
*
* 以123为例，在递归还没有开始前，先把空集push到result中，之后：
* 1. 以1位head，对23做dfs，所以pos需要加1，用于分支限界(1 12 13 123)
* 2. 回溯到1，以2为head，对3做dfs (2 23)
* 3. 回溯到3，以3为head，之后循环结束。 (3)
*
*
*/
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
// ensure that elements in a subset must be in non-descending order.
sort(nums.begin(), nums.end());
vector<vector<int>> res;
vector<int> path;
dfs(nums, res, path, 0);
return res;
}
private:
void dfs(const vector<int> &nums, vector<vector<int>> &res, vector<int> &path, int pos) {
res.push_back(path);
for (int i = pos; i < nums.size(); i++) {
path.push_back(nums[i]);
dfs(nums, res, path, i + 1);
path.pop_back();
}
}
};
https://leetcode.com/problems/subsets-ii/
Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note:
- Elements in a subset must be in non-descending order.
- The solution set must not contain duplicate subsets.
For example,
If nums =[1,2,2], a solution is:[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
同一位置上，前面取过的数，后面就不要再重复取了，当然当i = pos时，这个数必然是第一次取。
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int> &nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> res;
vector<int> path;
dfs(nums, res, path, 0);
return res;
}
private:
void dfs(const vector<int> &nums, vector<vector<int>> &res, vector<int> &path, int pos) {
res.push_back(path);
for (int i = pos; i < nums.size(); i++) {
if (i != pos && nums[i] == nums[i - 1]) {
continue;
}
path.push_back(nums[i]);
dfs(nums, res, path, i + 1);
path.pop_back();
}
}
};
（未完待续）
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原文地址：http://www.cnblogs.com/jianxinzhou/p/4712148.html
