【重点】【回溯】39.组合总和

题目

Python

法1：DFS+回溯_选或不选

# 写法1
class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res, path = [], []
        self.dfs(candidates, 0, res, path, target)

        return res

    def dfs(self, candidates, start, res, path, residual):
        if start <= len(candidates) and residual == 0:
            res.append(path.copy())
            return 
        
        if start >= len(candidates) or residual < 0:
            return
        
        path.append(candidates[start])
        self.dfs(candidates, start, res, path, residual - candidates[start]) # 选择candidates[start]
        path.pop()
        self.dfs(candidates, start+1, res, path, residual) # 不选择candidates[start]


# 写法2
class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res, tmp = list(), list()
        candidates.sort()
        self.dfs(candidates, 0, target, res, tmp)

        return res
    
    def dfs(self, candidates, start, target, res, tmp):
        if target == 0:
            res.append(tmp.copy())
            return 
        if start == len(candidates) or target < candidates[start]:
            return 

        self.dfs(candidates, start+1, target, res, tmp) # 不选择当前数字
        tmp.append(candidates[start])
        self.dfs(candidates, start, target - candidates[start], res, tmp) # 选择当前数字
        tmp.pop() # 回溯: 恢复现场

        # return # 最后这里其实有个默认return, 可写可不写!

旧解法

法1：回溯

必须掌握基本算法！！！

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        dfs(candidates, 0, target, res, tmp);

        return res;
    }

    public void dfs(int[] candidates, int idx, int target, List<List<Integer>> res, List<Integer> tmp) {
        if (idx == candidates.length) {
            return;
        }
        if (target == 0) {
            res.add(new ArrayList<>(tmp));
            return;
        }
        dfs(candidates, idx + 1, target, res, tmp); // 不选idx
        if (target >= candidates[idx]) {            // 选择idx
            tmp.add(candidates[idx]);
            dfs(candidates, idx, target - candidates[idx], res, tmp);
            tmp.remove(tmp.size() - 1);
        }
    }
}

法2：回溯+剪枝

先排序，再dfs，可以进一步剪枝
参考链接

class Solution {
    void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
        // 子集和等于 target 时，记录解
        if (target == 0) {
            res.add(new ArrayList<>(state));
            return;
        }
        // 遍历所有选择
        // 剪枝二：从 start 开始遍历，避免生成重复子集
        for (int i = start; i < choices.length; i++) {
            // 剪枝一：若子集和超过 target ，则直接结束循环
            // 这是因为数组已排序，后边元素更大，子集和一定超过 target
            if (target - choices[i] < 0) {
                break;
            }
            // 尝试：做出选择，更新 target, start
            state.add(choices[i]);
            // 进行下一轮选择
            backtrack(state, target - choices[i], choices, i, res);
            // 回退：撤销选择，恢复到之前的状态
            state.remove(state.size() - 1);
        }
    }

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<Integer> state = new ArrayList<>(); // 状态（子集）
        Arrays.sort(candidates); // 对 candidates 进行排序
        int start = 0; // 遍历起始点
        List<List<Integer>> res = new ArrayList<>(); // 结果列表（子集列表）
        backtrack(state, target, candidates, start, res);
        return res;
    }
}