「代码随想录算法训练营」第二十天 | 回溯算法 part2
39. 组合总和
题目链接:https://leetcode.cn/problems/combination-sum/
题目难度:中等
文章讲解:https://programmercarl.com/0039.组合总和.html
视频讲解:https://www.bilibili.com/video/BV1KT4y1M7HJ
题目状态:久违的通过!
思路:
使用回溯模板,在单层循环时判断当前数组值是否大于目标值,若大于,直接跳过,其他思路就是和回溯模板中一样,直接看代码。
代码:
class Solution {
public:
vector<vector<int>> res;
vector<int> vec;
void backtracking(vector<int> &candidates, int target, int startIndex) {
if(target == 0) {
res.push_back(vec);
return;
}
for(int i = startIndex; i < candidates.size(); ++i) {
if(candidates[i] > target)
continue;
vec.push_back(candidates[i]);
target -= candidates[i];
backtracking(candidates, target, i);
vec.pop_back();
target += candidates[i];
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
backtracking(candidates, target, 0);
return res;
}
};
剪枝优化后的代码:
class Solution {
private:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& candidates, int target, int sum, int startIndex) {
if (sum == target) {
result.push_back(path);
return;
}
// 如果 sum + candidates[i] > target 就终止遍历
for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
sum += candidates[i];
path.push_back(candidates[i]);
backtracking(candidates, target, sum, i);
sum -= candidates[i];
path.pop_back();
}
}
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
result.clear();
path.clear();
sort(candidates.begin(), candidates.end()); // 需要排序
backtracking(candidates, target, 0, 0);
return result;
}
};
40. 组合总和II
题目链接:https://leetcode.cn/problems/combination-sum-ii/
题目难度:中等
文章讲解:https://programmercarl.com/0040.组合总和II.html
视频讲解:https://www.bilibili.com/video/BV12V4y1V73A
题目状态:借助ChatGPT通过
思路:
在每层递归的时候,要确保当前递归的值和前一个递归的值不相等。因此需要首先对数组排序,之后在执行回溯三部曲。
代码:
class Solution {
public:
vector<vector<int>> res;
vector<int> vec;
void backtracking(vector<int> &candidates, int target, int startIdx) {
if(target == 0) {
res.push_back(vec);
return;
}
for(int i = startIdx; i < candidates.size(); ++i) {
if(i > startIdx && candidates[i] == candidates[i - 1])
continue;
if(target < candidates[i])
break;
target -= candidates[i];
vec.push_back(candidates[i]);
backtracking(candidates, target, i + 1);
target += candidates[i];
vec.pop_back();
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
backtracking(candidates, target, 0);
return res;
}
};
131. 分割回文串
题目链接:https://leetcode.cn/problems/palindrome-partitioning/
题目难度:中等
文章讲解:https://programmercarl.com/0131.分割回文串.html
视频讲解:https://www.bilibili.com/video/BV1c54y1e7k6
题目状态:这题好难,看题解通过
思路:
通过回溯,每层分割是按照几个元素来分,每次分割后判断分割出来的元素是否是一个回文串,直接理解代码会更好理解一些。
代码:
class Solution {
public:
vector<vector<string>> res;
vector<string> vec;
bool isLoop(string s, int start, int end) {
for(int i = start, j = end; i < j; i++, j--)
if(s[i] != s[j]) return false;
return true;
}
void backtracking(string s, int startIdx) {
if(startIdx >= s.size()) {
res.push_back(vec);
return;
}
for(int i = startIdx; i < s.size(); ++i) {
if(isLoop(s, startIdx, i)) {
string str = s.substr(startIdx, i - startIdx + 1);
vec.push_back(str);
} else {
continue;
}
backtracking(s, i + 1);
vec.pop_back();
}
}
vector<vector<string>> partition(string s) {
backtracking(s, 0);
return res;
}
};