[面试] 回溯题目小结

[Acceml]

每天发一道 leetcode 题目，最近发现 A 的回溯题目有点多，基本就用一个模板搞定。

• [ Leetcode ] 77. 组合
• [ Leetcode ] 60. 第 k 个排列
• [ Leetcode ] 47. 全排列 II
• [ Leetcode ] 46.全排列

[ Leetcode ] 79. 子集

java

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int[] nums, int start) {
        list.add(new ArrayList<>(tempList));
        for (int i = start; i < nums.length; i++) {
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.remove(tempList.size() - 1);
        }
    }
}

python

class Solution:
    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        list = []
        nums.sort()
        self.bracktrack(list, [], nums, 0)
        return list

    def bracktrack(self, list, tempList, nums, start):
        list.append(tempList.copy())
        for i in range(start, len(nums)):
            tempList.append(nums[i])
            self.bracktrack(list, tempList, nums, i + 1)
            tempList.pop()

[ Leetcode ] 77. 组合

java 版本

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> res = new ArrayList<>();
        backtrack(res, n, 1, k, new ArrayList<>());
        return res;
    }

    public void backtrack(List<List<Integer>> res, int n, int num, int k, List<Integer> list) {
        if (list.size() == k) {
            res.add(new ArrayList<>(list));
        } else {
            for (int i = num; i <= n; i++) {
                list.add(i);
                backtrack(res, n, i + 1, k, list);
                list.remove(list.size() - 1);
            }
        }
    }
}

python 版本

class Solution:
    def backtrack(self, res, n, nums, k, current):
        if len(current) == k:
            res.append(current.copy())
        else:
            for i in range(nums, n + 1):
                current.append(i)
                self.backtrack(res, n, i + 1, k, current)
                current.pop()

    def combine(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: List[List[int]]
        """
        res = []
        self.backtrack(res, n, 1, k, [])
        return res

[ Leetcode ] 46.全排列

class Solution {
    public static List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        backtrack(res, new ArrayList<>(), nums);
        return res;
    }

    public static void backtrack(List<List<Integer>> res, List<Integer> tempRes, int[] nums) {
        if (tempRes.size() == nums.length) {
            //遍历完了,不需要回溯了.
            res.add(new ArrayList<>(tempRes));
        } else {
            for (int i = 0; i < nums.length; i++) {
                if (tempRes.contains(nums[i])) {
                    //System.out.println("i:" + i + "---" + tempRes.stream().map(String::valueOf).collect(Collectors.joining(",")));
                    continue;
                }
                tempRes.add(nums[i]);
                //System.out.println("i:" + i + "---" + tempRes.stream().map(String::valueOf).collect(Collectors.joining(",")));
                backtrack(res, tempRes, nums);
                tempRes.remove(tempRes.size() - 1);
                //System.out.println("i:" + i + ":afterRemoveLast---" + tempRes.stream().map(String::valueOf).collect(Collectors.joining(",")));
            }
        }
    }
}

[ Leetcode ] 47. 全排列 II

class Solution {
    public static List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        boolean[] used = new boolean[nums.length];
        backtrack(res, new ArrayList<>(), nums, used);
        return res;
    }

    public static void backtrack(List<List<Integer>> res, List<Integer> tempRes, int[] nums, boolean[] used) {
        if (tempRes.size() == nums.length) {
            //遍历完了,不需要回溯了.
            res.add(new ArrayList<>(tempRes));
        } else {
            for (int i = 0; i < nums.length; i++) {
                if (used[i]) {
                    continue;
                }

                if (i > 0 && nums[i - 1] == nums[i] && !used[i - 1]) {
                    continue;
                }

                used[i] = true;
                tempRes.add(nums[i]);
                backtrack(res, tempRes, nums, used);
                used[i] = false;
                tempRes.remove(tempRes.size() - 1);
            }
        }
    }
}