leetcode40. 组合总和 II

1.题目描述：

给定一个候选人编号的集合candidates和一个目标数target，找出candidates中所有可以使数字和为target的组合。candidates中的每个数字在每个组合中只能使用一次。注意：解集不能包含重复的组合。

2.当开始写的回溯代码（已完成剪枝）：

类似的题型做多了直接按照模板写。一跑发现有重复的结果。原因是此题数组元素会有重复的，[10,1,2,7,6,1,5]排序后[1,1,2,5,6,7,10]，[1,2,5]和[1,7]会出现两次，这里的1两次分别是索引0和索引1的位置，可以在最后的resList中去重但是太麻烦。对数组排序之后，去重只要找到元素相等的地方。会发现[1,1,6]只有一次，也就是纵向/递归往下/树枝上允许出现重复元素，但是横线/同一树层上不允许出现重复，否则就会有我上述跑完结果（同一for循环中若有相同元素，那么符合条件的话必然也是两份）。

class Solution {
    private List> resList = new ArrayList<>();
    private List list = new ArrayList<>();
    private int sum = 0;
    public List> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        backTracking(candidates, target, 0);
        return resList;
    }

    public void backTracking(int[] candidates, int target, int index) {
        if (sum == target) {
            resList.add(new ArrayList<>(list));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            if (sum + candidates[i] > target) break;//剪枝
            list.add(candidates[i]);
            sum += candidates[i];
            backTracking(candidates, target, i + 1);
            list.remove(list.size() - 1);
            sum -= candidates[i];
        }
    }
}

3.利用index去重（推荐）：

class Solution {
    private List> resList = new ArrayList<>();
    private List list = new ArrayList<>();
    private int sum = 0;
    public List> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        backTracking(candidates, target, 0);
        return resList;
    }

    public void backTracking(int[] candidates, int target, int index) {
        if (sum == target) {
            resList.add(new ArrayList<>(list));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            if (i > index && candidates[i] == candidates[i - 1]) continue;//去重，i > index，防止空指针也防止[1,1,6]放不进结果集，不能i > 0开始；这里操作candidates[i - 1]树枝时已经把一份结果全存入
            if (sum + candidates[i] > target) break;//剪枝
            list.add(candidates[i]);
            sum += candidates[i];
            backTracking(candidates, target, i + 1);
            list.remove(list.size() - 1);
            sum -= candidates[i];
        }
    }
}

二刷：

class Solution {
    List> res = new ArrayList<>();
    List list = new ArrayList<>();
    public List> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        backTracking(candidates, target, 0);
        return res;
    }
    public void backTracking(int[] candidates, int target, int index) {
        int sum = 0;
        for (Integer item : list) sum += item;
        if (sum == target) res.add(new ArrayList<>(list));
        if (sum >= target) return;
        for (int i = index; i < candidates.length && sum + candidates[i] <= target; i++) {
            if (i > index && candidates[i] == candidates[i - 1]) continue;
            list.add(candidates[i]);
            backTracking(candidates, target, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

4.使用标记数组去重：

class Solution {
    private List> resList = new ArrayList<>();
    private List list = new ArrayList<>();
    private int sum = 0;
    private boolean[] flag;
    public List> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        flag = new boolean[candidates.length];//标记数组帮助判断同一树层元素是否已经被访问
        backTracking(candidates, target, 0);
        return resList;
    }

    public void backTracking(int[] candidates, int target, int index) {
        if (sum == target) {
            resList.add(new ArrayList<>(list));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            if (i > 0 && candidates[i] == candidates[i - 1] && !flag[i - 1]) continue;//去重，这里i > 0没问题，因为有标记的判断
            //flag[i - 1] == true，说明纵向同一树枝candidates[i - 1]使用过
            //flag[i - 1] == false，说明横向同一树层candidates[i - 1]使用过，但是被回溯复原
            if (sum + candidates[i] > target) break;//剪枝
            list.add(candidates[i]);
            flag[i] = true;
            sum += candidates[i];
            backTracking(candidates, target, i + 1);
            list.remove(list.size() - 1);//回溯
            sum -= candidates[i];//回溯
            flag[i] = false;//回溯
        }
    }
}

5.另一种使用标记数组去重（推荐）：

上一种标记数组记录了位置而非值（排序后的原因），这种依据每个位置的值（无需排序，排序为了剪枝）来去重，和leetcode491. 递增子序列去重方法一致，但是子集问题必须先排序（不排序会出现虽是同一树层上取不重复值的情况，担仍出现重复子集）！

class Solution {
    private List> resList = new ArrayList<>();
    private List list = new ArrayList<>();
    private int sum = 0;
    public List> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        backTracking(candidates, target, 0);
        return resList;
    }
 
    public void backTracking(int[] candidates, int target, int index) {
        if (sum == target) {
            resList.add(new ArrayList<>(list));
            return;
        }
        boolean[] flag = new boolean[51];//每层更新
        for (int i = index; i < candidates.length; i++) {
            if (flag[candidates[i]]) continue;//去重
            if (sum + candidates[i] > target) break;//剪枝
            list.add(candidates[i]);
            flag[candidates[i]] = true;//无需回溯
            sum += candidates[i];
            backTracking(candidates, target, i + 1);
            list.remove(list.size() - 1);//回溯
            sum -= candidates[i];//回溯
        }
    }
}