491.递增子序列
cpp
class Solution {
private:
vector<vector<int>> res;
vector<int> temp;
void backtracing (vector<int>& nums, int startIndex) {
if (temp.size() >= 2) {
res.push_back(temp);
//return;
}
unordered_set<int> uset; // 使用set来对本层元素进行去重
for (int i = startIndex; i < nums.size(); ++i) {
if ((!temp.empty() && nums[i] < temp.back()) || uset.find(nums[i]) != uset.end()) {
continue;
}
uset.insert(nums[i]); // 记录这个元素在本层用过了，本层后面不能再用了
temp.push_back(nums[i]);
backtracing(nums, i + 1);
temp.pop_back();
}
}
public:
vector<vector<int>> findSubsequences(vector<int>& nums) {
backtracing(nums, 0);
return res;
}
};46.全排列
cpp
class Solution {
private:
vector<vector<int>> res;
vector<int> temp;
void backtracing (vector<int>& nums) {
if (temp.size() == nums.size()) {
res.push_back(temp);
return;
}
for (int i = 0; i < nums.size(); ++i) {
if (find(temp.begin(), temp.end(), nums[i]) != temp.end()) { // nums 不含重复元素，可以直接判断元素是否被使用
continue;
}
temp.push_back(nums[i]);
backtracing(nums);
temp.pop_back();
}
}
public:
vector<vector<int>> permute(vector<int>& nums) {
backtracing(nums);
return res;
}
};使用used数组存储元素的使用状态
cpp
class Solution {
public:
vector<vector<int>> res;
vector<int> temp;
void backtracking(vector<int>& nums, vector<bool>& used){
if(temp.size() == nums.size()){
res.push_back(temp);
return;
}
//nums中不含重复元素
for(int i = 0; i < nums.size(); ++i){
if (used[i] == true) continue; // temp里已经收录的元素，直接跳过
used[i] = true;
temp.push_back(nums[i]);
backtracking(nums,used);
temp.pop_back();
used[i] = false;
}
}
vector<vector<int>> permute(vector<int>& nums) {
//使用used数组来记录nums中的每一位是否被使用
vector<bool> used(nums.size(), false);
backtracking(nums,used);
return res;
}
};47.全排列 II
cpp
class Solution {
private:
vector<vector<int>> res;
vector<int> temp;
void backtracing(vector<int>& nums, vector<bool>& used) {
if (temp.size() == nums.size()) {
res.push_back(temp);
return;
}
for (int i = 0; i < nums.size(); ++i) {
if((used[i] == true) || (i > 0 && used[i - 1] == false && nums[i] == nums[i - 1])) {
// 同一树杈的同一元素 已被使用过
// 同一树层 存在元素相等的情况
continue;
}
used[i] = true;
temp.push_back(nums[i]);
backtracing(nums, used);
temp.pop_back();
used[i] = false;
}
}
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<bool> used(nums.size(), false);
backtracing(nums, used);
return res;
}
};