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216.组合总和III

题目链接: 组合总和III

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class Solution {

    List<Integer> path = new ArrayList<>();
    List<List<Integer>> result = new ArrayList<>();

    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(n,0,k,1);
        return result;
    }

    public void backtracking(int targetSum,int sum,int k,int startIndex) {
        if(sum > targetSum){
            return ;
        }

        if(path.size() == k) {
            if(sum == targetSum){
                result.add(new ArrayList<>(path));
                return ;
            }
        }

        for(int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {
            path.add(i);
            sum += i;
            backtracking(targetSum,sum,k,i + 1);
            path.remove(path.size() - 1);
            sum -= i;
        }
    }
}

17.电话号码的字母组合

题目链接: 电话号码的字母组合

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class Solution {

    //设置全局列表存储最后的结果
    List<String> list = new ArrayList<>();

    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) {
            return list;
        }
        //初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""
        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        //迭代处理
        backTracking(digits, numString, 0);
        return list;

    }

    //每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuild
    StringBuilder temp = new StringBuilder();

    //比如digits如果为"23",num 为0，则str表示2对应的 abc
    public void backTracking(String digits, String[] numString, int num) {
        //遍历全部一次记录一次得到的字符串
        if (num == digits.length()) {
            list.add(temp.toString());
            return;
        }
        //str 表示当前num对应的字符串
        String str = numString[digits.charAt(num) - '0'];
        for (int i = 0; i < str.length(); i++) {
            temp.append(str.charAt(i));
            //c
            backTracking(digits, numString, num + 1);
            //剔除末尾的继续尝试
            temp.deleteCharAt(temp.length() - 1);
        }
    }
}