力扣216题、39题、40题（组合总和）

216、组合总和III

基本思想：

回溯法

具体实现：

剪枝优化：

1.元素总和大于目标值，在递归终止的地方剪枝

2.和77题思路一样，for循环的范围剪枝

代码：

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backTracking(n, k, 1, 0);
        return result;
    }
    private void backTracking(int targrtSum, int k, int startIndex, int sum){
        if(sum > targrtSum){
            return;//剪枝
        }
        if (path.size() == k){
            if (sum == targrtSum) {
                result.add(new ArrayList<>(path));
            }
            return;
        }
        for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++){
            path.add(i);
            sum += i;
            backTracking(targrtSum,k,i+1,sum);
            path.removeLast();
            sum -= i;
        }
    }
}

 

 

39、组合总和

具体实现：

 

 

代码：

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> track = new LinkedList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        traversal(candidates, target, 0, 0);
        return result;
    }
    public void traversal(int[] candidates, int target, int sum, int idx) {
        if (sum > target) return;
        if (sum == target) {
            result.add(new ArrayList(track));
            return;
        }
        for (int i = idx; i < candidates.length; i++) {
            sum += candidates[i];
            track.add(candidates[i]);
            traversal(candidates, target, sum, i);
            sum -= candidates[i];
            track.removeLast();
        }
    }
}

 

剪枝：

先对candidates进行从小到大的排序

本层的sum+candidates[i] >traget,就跳出循环

代码：

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> track = new LinkedList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        traversal(candidates, target, 0, 0);
        return result;
    }
    public void traversal(int[] candidates, int target, int sum, int idx) {
        if (sum == target) {
            result.add(new ArrayList(track));
            return;
        }
        for (int i = idx; i < candidates.length; i++) {
            if (sum + candidates[i] > target) break;
            sum += candidates[i];
            track.add(candidates[i]);
            traversal(candidates, target, sum, i);
            sum -= candidates[i];
            track.removeLast();
        }
    }
}

 

40、组合总和II

基本思想：

与39.组合总和套路相同，此题还需要加一个bool型数组used，用来记录同一树枝上的元素是否使用过。

具体实现：

使用数组中相同的元素有两种场景，

一种是同一树枝上，这个允许相同

另一种是同一树层上，这个不允许相同，此时需要跳出这一循环，进入下一循环

 

如果condidate[i]=condidate[i-1],判断是同一树枝的元素相等还是同一树层的元素相同？

1.用used数组，

• used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
• used[i - 1] == false，说明同一树层candidates[i - 1]使用过

2.直接跳过同一树层中使用过的元素

代码：

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> track = new LinkedList<>();
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        boolean[] used = new boolean[candidates.length];
        traversal(candidates, target, 0, 0, used);
        return result;
    }
    public void traversal(int[] candidates, int target, int sum, int idx, boolean[] used) {
        
        if (sum == target) {
            result.add(new ArrayList(track));
            return;
        }
        for (int i = idx; i < candidates.length; i++) {
            if (i > 0 && candidates[i] == candidates[i-1] && used[i-1] == false) continue;
            if (sum + candidates[i] > target) break;
            sum += candidates[i];
            track.add(candidates[i]);
            used[i] = true;
            traversal(candidates, target, sum, i + 1, used);
            sum -= candidates[i];
            track.removeLast();
            used[i] = false;
        }
    }
}

 

 

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> track = new LinkedList<>();
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        traversal(candidates, target, 0, 0);
        return result;
    }
    public void traversal(int[] candidates, int target, int sum, int idx) {
        
        if (sum == target) {
            result.add(new ArrayList(track));
            return;
        }
        for (int i = idx; i < candidates.length; i++) {
            //跳过同一树层使用过的元素
            if (i > idx && candidates[i] == candidates[i-1]) continue;
            if (sum + candidates[i] > target) break;
            sum += candidates[i];
            track.add(candidates[i]);
            traversal(candidates, target, sum, i + 1);
            sum -= candidates[i];
            track.removeLast();
        }
    }
}