Backtracking Algorithm Spike 2

Elements are weightless and non-checkable

Subset

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    // 记录回溯算法的递归路径
    LinkedList<Integer> track = new LinkedList<>();

    // 主函数
    public List<List<Integer>> subsets(int[] nums) {
        backtrack(nums, 0);
        return res;
    }

    // 回溯算法核心函数，遍历子集问题的回溯树
    void backtrack(int[] nums, int start) {

        // 前序位置，每个节点的值都是一个子集
        res.add(new LinkedList<>(track));
        
        // 回溯算法标准框架
        for (int i = start; i < nums.length; i++) {
            // 做选择
            track.addLast(nums[i]);
            // 通过 start 参数控制树枝的遍历，避免产生重复的子集
            backtrack(nums, i + 1);
            // 撤销选择
            track.removeLast();
        }
    }
}

combination 

Given two integers  n and  sum k, return  [1, n] all possible  k combinations of numbers in the range.

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    // 记录回溯算法的递归路径
    LinkedList<Integer> track = new LinkedList<>();

    // 主函数
    public List<List<Integer>> combine(int n, int k) {
        backtrack(1, n, k);
        return res;
    }

    void backtrack(int start, int n, int k) {
        // base case
        if (k == track.size()) {
            // 遍历到了第 k 层，收集当前节点的值
            res.add(new LinkedList<>(track));
            return;
        }
        
        // 回溯算法标准框架
        for (int i = start; i <= n; i++) {
            // 选择
            track.addLast(i);
            // 通过 start 参数控制树枝的遍历，避免产生重复的子集
            backtrack(i + 1, n, k);
            // 撤销选择
            track.removeLast();
        }
    }
}

full array 

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    // 记录回溯算法的递归路径
    LinkedList<Integer> track = new LinkedList<>();
    // track 中的元素会被标记为 true
    boolean[] used;

    /* 主函数，输入一组不重复的数字，返回它们的全排列 */
    public List<List<Integer>> permute(int[] nums) {
        used = new boolean[nums.length];
        backtrack(nums);
        return res;
    }

    // 回溯算法核心函数
    void backtrack(int[] nums) {
        // base case，到达叶子节点
        if (track.size() == nums.length) {
            // 收集叶子节点上的值
            res.add(new LinkedList(track));
            return;
        }

        // 回溯算法标准框架
        for (int i = 0; i < nums.length; i++) {
            // 已经存在 track 中的元素，不能重复选择
            if (used[i]) {
                continue;
            }
            // 做选择
            used[i] = true;
            track.addLast(nums[i]);
            // 进入下一层回溯树
            backtrack(nums);
            // 取消选择
            track.removeLast();
            used[i] = false;
        }
    }
}

In summary, if it is a subset/combination problem, the method is set to backtrack(int[] nums,int start) , and backtrack(nums,i+1)    is used for recursion . However, if it is a permutation problem, the method is set to backtrack (int[] nums) and additionally set a used array to mark whether it has been selected , and use backtrack(nums) when recursive

Elements can be repeated but not reselected

Elements may contain repeated elements, but they can only be selected once

Subset 

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> track = new LinkedList<>();

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        // 先排序，让相同的元素靠在一起
        Arrays.sort(nums);
        backtrack(nums, 0);
        return res;
    }

    void backtrack(int[] nums, int start) {
        // 前序位置，每个节点的值都是一个子集
        res.add(new LinkedList<>(track));
        
        for (int i = start; i < nums.length; i++) {
            // 剪枝逻辑，值相同的相邻树枝，只遍历第一条
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }
            track.addLast(nums[i]);
            backtrack(nums, i + 1);
            track.removeLast();
        }
    }
}

Compared with the previous subsets, here is an advanced array sorting, and a pruning link is added during the loop 

combination 

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    // 记录回溯的路径
    LinkedList<Integer> track = new LinkedList<>();
    // 记录 track 中的元素之和
    int trackSum = 0;

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        if (candidates.length == 0) {
            return res;
        }
        // 先排序，让相同的元素靠在一起
        Arrays.sort(candidates);
        backtrack(candidates, 0, target);
        return res;
    }

    // 回溯算法主函数
    void backtrack(int[] nums, int start, int target) {
        // base case，达到目标和，找到符合条件的组合
        if (trackSum == target) {
            res.add(new LinkedList<>(track));
            return;
        }
        // base case，超过目标和，直接结束
        if (trackSum > target) {
            return;
        }

        // 回溯算法标准框架
        for (int i = start; i < nums.length; i++) {
            // 剪枝逻辑，值相同的树枝，只遍历第一条
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }
            // 做选择
            track.add(nums[i]);
            trackSum += nums[i];
            // 递归遍历下一层回溯树
            backtrack(nums, i + 1, target);
            // 撤销选择
            track.removeLast();
            trackSum -= nums[i];
        }
    }
}

Compared with the previous combination, the advanced array sorting is added here, and the pruning link is added during the loop 

full array

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> track = new LinkedList<>();
    boolean[] used;

    public List<List<Integer>> permuteUnique(int[] nums) {
        // 先排序，让相同的元素靠在一起
        Arrays.sort(nums);
        used = new boolean[nums.length];
        backtrack(nums);
        return res;
    }

    void backtrack(int[] nums) {
        if (track.size() == nums.length) {
            res.add(new LinkedList(track));
            return;
        }

        for (int i = 0; i < nums.length; i++) {
            if (used[i]) {
                continue;
            }
            // 新添加的剪枝逻辑，固定相同的元素在排列中的相对位置
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
                continue;
            }
            track.add(nums[i]);
            used[i] = true;
            backtrack(nums);
            track.removeLast();
            used[i] = false;
        }
    }
}

 Compared with the previous full arrangement, here is an advanced array sorting, and a pruning link is added during the loop ( in particular, the pruning condition needs to be added !used[i-1] ) 

Elements are not repeated and can be selected 

There are no repeated elements in the array, but you can choose repeatedly 

combination 

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    // 记录回溯的路径
    LinkedList<Integer> track = new LinkedList<>();
    // 记录 track 中的路径和
    int trackSum = 0;

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates.length == 0) {
            return res;
        }
        backtrack(candidates, 0, target);
        return res;
    }

    // 回溯算法主函数
    void backtrack(int[] nums, int start, int target) {
        // base case，找到目标和，记录结果
        if (trackSum == target) {
            res.add(new LinkedList<>(track));
            return;
        }
        // base case，超过目标和，停止向下遍历
        if (trackSum > target) {
            return;
        }

        // 回溯算法标准框架
        for (int i = start; i < nums.length; i++) {
            // 选择 nums[i]
            trackSum += nums[i];
            track.add(nums[i]);
            // 递归遍历下一层回溯树
            // 同一元素可重复使用，注意参数
            backtrack(nums, i, target);
            // 撤销选择 nums[i]
            trackSum -= nums[i];
            track.removeLast();
        }
    }
}

Compared with elements that can be repeated but not reselected, just change a sentence when looping backtrack(nums, i, target); 

full array

class Solution {

    List<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> track = new LinkedList<>();

    public List<List<Integer>> permuteRepeat(int[] nums) {
        backtrack(nums);
        return res;
    }

    // 回溯算法核心函数
    void backtrack(int[] nums) {
        // base case，到达叶子节点
        if (track.size() == nums.length) {
            // 收集叶子节点上的值
            res.add(new LinkedList(track));
            return;
        }

        // 回溯算法标准框架
        for (int i = 0; i < nums.length; i++) {
            // 做选择
            track.add(nums[i]);
            // 进入下一层回溯树
            backtrack(nums);
            // 取消选择
            track.removeLast();
        }
    }
}

Compared with the full arrangement in the above two cases, there is no need to add the used array to mark whether it has been used and the pruning link and the array in advance