算法训练day29 LeetCode 39.40.131

39.组合总和

题目

39. 组合总和 - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

• class Solution
  {
  private:
      vector<vector<int>> result;
      vector<int> path;
      void backtracking(vector<int> &candidates, int target, int sum, int startIndex)
      {
          if (sum > target)
              return;
          if (sum == target)
          {
              result.push_back(path);
          }
          for (int i = startIndex; i < candidates.size(); 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();
          backtracking(candidates, target, 0, 0);
          return result;
      }
  };
  

• 在sum>target时，依然会进入下一层循环，所以可以添加一个剪枝

•     void backtracking(vector<int> &candidates, int target, int sum, int startIndex)
      {
         // if (sum > target)
         //     return;
          if (sum == target)
          {
              result.push_back(path);
              return;
          }
          for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) //剪枝
          {
              sum += candidates[i];
              path.push_back(candidates[i]);
              backtracking(candidates, target, sum, i);
              sum -= candidates[i];
              path.pop_back();
          }
      }
  

40.组合总和II

题目

40. 组合总和 II - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

• candidates 中元素重复、单个结果中元素可以重复、总结果集中不能有结果重复

• 需要去重 ---> 将组合的形式以树形结构组织

  • 单个结果元素可重复 ---> 单次组合（单枝）中不需要去重
  • 多个结果之间不能重复 ---> 不同次组合（同一层）中需要去重 ---> 将元素排序之后组合实现去重

• class Solution
  {
  private:
      vector<vector<int>> result;
      vector<int> path;
      void backtracking(vector<int> &candidates, int target, int sum, int startIndex, vector<bool> used)
      {
          if (sum == target)
          {
              result.push_back(path);
              return;
          }
          for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++)
          {
              if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) //同层去重
                  continue;
              sum += candidates[i];
              path.push_back(candidates[i]);
              used[i] = true;
              backtracking(candidates, target, sum, i + 1, used);
              used[i] = false;
              sum -= candidates[i];
              path.pop_back();
          }
      }
  
  public:
      vector<vector<int>> combinationSum2(vector<int> &candidates, int target)
      {
          vector<bool> used(candidates.size(), false);
          result.clear();
          path.clear();
          sort(candidates.begin(), candidates.end());
          backtracking(candidates, target, 0, 0, used);
          return result;
      }
  };
  

131.分割回文串

题目

131. 分割回文串 - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

• 递归用来纵向遍历，for循环用来横向遍历，切割线（就是图中的红线）切割到字符串的结尾位置，说明找到了一个切割方法。

• class Solution
  {
  private:
      vector<vector<string>> result;
      vector<string> path;
      bool isPalindrome(const 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(const string &s, int startIndex)
      {
          if (startIndex >= s.size())
          {
              result.push_back(path);
              return;
          }
          for (int i = startIndex; i < s.size(); i++)
          {
              if (isPalindrome(s, startIndex, i))
              {
                  string str = s.substr(startIndex, i - startIndex + 1);
                  path.push_back(str);
              }
              else
                  continue;
              backtracking(s, i + 1);
              path.pop_back();
          }
      }
  
  public:
      vector<vector<string>> partition(string s)
      {
          result.clear();
          path.clear();
          backtracking(s, 0);
          return result;
      }
  };