[Leetcode][第39题][JAVA][组合总和][回溯][dfs][剪枝]

向右看齐 2022-12-05 05:11 182阅读 0赞

#### 【问题描述】\[中等\] ####

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center]

#### 【解答思路】 ####

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 2]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 3]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 4]

###### 1. 回溯 ######

import java.util.ArrayDeque;
    import java.util.ArrayList;
    import java.util.Deque;
    import java.util.List;
    
    public class Solution { 
    
        public List<List<Integer>> combinationSum(int[] candidates, int target) { 
            int len = candidates.length;
            List<List<Integer>> res = new ArrayList<>();
            if (len == 0) { 
                return res;
            }
    
            Deque<Integer> path = new ArrayDeque<>();
            dfs(candidates, 0, len, target, path, res);
            return res;
        }
    
        /** * @param candidates 候选数组 * @param begin 搜索起点 * @param len 冗余变量，是 candidates 里的属性，可以不传 * @param target 每减去一个元素，目标值变小 * @param path 从根结点到叶子结点的路径，是一个栈 * @param res 结果集列表 */
        private void dfs(int[] candidates, int begin, int len, int target, Deque<Integer> path, List<List<Integer>> res) { 
            // target 为负数和 0 的时候不再产生新的孩子结点
            if (target < 0) { 
                return;
            }
            if (target == 0) { 
                res.add(new ArrayList<>(path));
                return;
            }
    
            // 重点理解这里从 begin 开始搜索的语意
            for (int i = begin; i < len; i++) { 
                path.addLast(candidates[i]);
    
                // 注意：由于每一个元素可以重复使用，下一轮搜索的起点依然是 i，这里非常容易弄错
                dfs(candidates, i, len, target - candidates[i], path, res);
    
                // 状态重置
                path.removeLast();
            }
        }
    }
    
    作者：liweiwei1419
    链接：https://leetcode-cn.com/problems/combination-sum/solution/hui-su-suan-fa-jian-zhi-python-dai-ma-java-dai-m-2/
    来源：力扣（LeetCode）
    著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

**正向思维 求和（增加了空间和变量）**

class Solution { 
        public List<List<Integer>> res = new ArrayList<>();
        public List<List<Integer>> combinationSum(int[] candidates, int target) { 
            dfs(candidates,target,new ArrayDeque<Integer>(),0,0);
            return res;
        }
        public void dfs(int[] cand,int target,Deque<Integer> temp,int sum,int index){ 
            if(sum>target) return;
            if(sum==target){ 
                res.add(new ArrayList<>(temp));
                return;
            }
            for(int i=index;i<cand.length;++i){ 
                temp.addLast(cand[i]);
                dfs(cand,target,temp,sum+cand[i],i);
                temp.removeLast();
            }
        }
    }

###### 2. 回溯剪树枝 ######

![在这里插入图片描述][20200909111235399.png_pic_center]

import java.util.ArrayDeque;
    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.Deque;
    import java.util.List;
    
    public class Solution { 
    
        public List<List<Integer>> combinationSum(int[] candidates, int target) { 
            int len = candidates.length;
            List<List<Integer>> res = new ArrayList<>();
            if (len == 0) { 
                return res;
            }
    
            // 排序是剪枝的前提
            Arrays.sort(candidates);
            Deque<Integer> path = new ArrayDeque<>();
            dfs(candidates, 0, len, target, path, res);
            return res;
        }
    
        private void dfs(int[] candidates, int begin, int len, int target, Deque<Integer> path, List<List<Integer>> res) { 
            // 由于进入更深层的时候，小于 0 的部分被剪枝，因此递归终止条件值只判断等于 0 的情况
            if (target == 0) { 
                res.add(new ArrayList<>(path));
                return;
            }
    
            for (int i = begin; i < len; i++) { 
                // 重点理解这里剪枝，前提是候选数组已经有序，
                if (target - candidates[i] < 0) { 
                    break;
                }
                
                path.addLast(candidates[i]);
                dfs(candidates, i, len, target - candidates[i], path, res);
                path.removeLast();
            }
        }
    }
    
    作者：liweiwei1419
    链接：https://leetcode-cn.com/problems/combination-sum/solution/hui-su-suan-fa-jian-zhi-python-dai-ma-java-dai-m-2/

#### 【总结】 ####

###### 1. 总结![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 5] ######

###### 2.在函数中调用dfs时，直接初始化变量，不用另外创建新的变量 ######

![在这里插入图片描述][2020090911180096.png_pic_center]

###### 3.回溯算法相关题目 ######

[\[Leedcode\]\[JAVA\]\[第46题\]\[全排列\]\[回溯算法\]][Leedcode_JAVA_46]  
[\[Leetcode\]\[第81题\]\[JAVA\]\[N皇后问题\]\[回溯算法\]][Leetcode_81_JAVA_N]  
[\[Leetcode\]\[第60题\]\[JAVA\]\[第k个排列\]\[回溯\]\[DFS\]\[剪枝\]][Leetcode_60_JAVA_k_DFS]

转载链接：https://leetcode-cn.com/problems/combination-sum/solution/hui-su-suan-fa-jian-zhi-python-dai-ma-java-dai-m-2/

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center]: /images/20221123/4735abf30db64ac8aef20655de19201c.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 1]: /images/20221123/3e3b10af3d264d25a955d9f7f95bbfc9.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 2]: /images/20221123/c98e5b3690a24798b6b6cbf17cbf1522.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 3]: /images/20221123/2f27e060b8c1414eb287b35ddc2fa664.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 4]: /images/20221123/0ae84587654841cd9060e7d9151bddc5.png
[20200909111235399.png_pic_center]: /images/20221123/177cf7900eb540b9a8864e76d3591274.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70_pic_center 5]: /images/20221123/d80724e173fd4c8199d93954a24e36c1.png
[2020090911180096.png_pic_center]: /images/20221123/3c48f6dfb5c2401398f0d8e5a15675ee.png
[Leedcode_JAVA_46]: https://blog.csdn.net/dadongwudi/article/details/105760561
[Leetcode_81_JAVA_N]: https://blog.csdn.net/dadongwudi/article/details/108379661
[Leetcode_60_JAVA_k_DFS]: https://blog.csdn.net/dadongwudi/article/details/108419966