代码随想录训练营第二十五天 ｜ 回溯算法

今天是第二十五天，回溯算法

216. 组合总和III

class Solution {
    List<List<Integer>> res = new ArrayList<>();

    List<Integer> paths = new ArrayList<>();
    int sum = 0;
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtrack(k, n, 1);
        return res;
    }
    public void backtrack(int k, int n, int startIndex){
        if(sum>n || paths.size()+n - startIndex + 1 < k){
            return;
        }
        if(sum == n && paths.size() == k){
            res.add(new ArrayList<>(paths));
            return;
        }
        for(int i = startIndex; i <= n && i <= 9; i++){
            paths.add(i);
            sum += i;
            backtrack(k, n, i + 1);
            sum -= i;
            paths.remove(paths.size() - 1);
        }
    }
}

很典型的回溯算法题，求1-9相加为n的k个个数组合

17. 电话号码的字母组合

class Solution {

    String[] mappings = new String[]{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    List<String> res = new ArrayList<>();

    public List<String> letterCombinations(String digits) {
        if (digits == "" || digits.length() == 0) {
            return res;
        }
        func(new StringBuilder(), digits, 0);
        return res;
    }

    private void func(StringBuilder sb, String digits, int index) {
        if (sb.length() == digits.length()) {
            res.add(sb.toString());
            return;
        }
        Integer num = Integer.valueOf(String.valueOf(digits.charAt(index)));
        for (int i = 0; i < mappings[num].length(); i++) {
            sb.append(mappings[num].charAt(i));
            func(sb, digits, index + 1);
            sb.deleteCharAt(index);
        }
    }
}

没想出来完全

回溯算法这块还是不是很理解 得练