【算法刷题day24】Leetcode：216. 组合总和 III、17. 电话号码的字母组合

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java的Deque

Leetcode 216. 组合总和 III

题目：216. 组合总和 III
解析：代码随想录解析

解题思路

回溯三部曲：确定递归函数参数；确定终止条件；单层搜索过程

代码

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> paths = new ArrayList<>();
    int curSum = 0;
    private void backtracking(int k, int n, int startIndex) {
    //忘了这个减枝
    	if (curSum > n) return;
        if (paths.size() == k) {
            if (curSum == n)
                res.add(new ArrayList<>(paths));
            return;
        }
        for (int i = startIndex; i <= 9; i++) {
            paths.add(i);
            curSum += i;
            backtracking(k, n, i + 1);
            paths.remove(paths.size()-1);
            curSum -= i;
        }
    }
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);
        return res;
    }
}

总结

暂无

Leetcode 17. 电话号码的字母组合

题目：17. 电话号码的字母组合
解析：代码随想录解析

解题思路

函数参数遍历数字，函数内遍历数字按键对应的字母

代码

class Solution {
    List<String> res = new ArrayList<>();
    StringBuilder sb = new StringBuilder();
    private void backtracking(String digits, String[] numString, int startIndex) {
        if (startIndex == digits.length()) {
            if (sb.length() > 0)
                res.add(sb.toString());
            return;
        }
        String tmp = numString[digits.charAt(startIndex) - '0'];
        for (int i = 0; i < tmp.length(); i++) {
            sb.append(tmp.charAt(i));
            backtracking(digits, numString, startIndex + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
    public List<String> letterCombinations(String digits) {
        String[] NUMSTRING = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs","tuv", "wxyz"};
        backtracking(digits, NUMSTRING, 0);
        return res;
    }
}

总结

暂无