Leetcode 17. Letter Combinations of a Phone Number

再见小小ronnie

于 2017-03-31 02:06:58 发布

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分类专栏： Leetcode DFS

本文链接：https://blog.youkuaiyun.com/u011094114/article/details/68754585

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本文介绍了一个使用递归方法实现的算法，该算法可以根据输入的数字字符串返回所有可能的字母组合，这些字母对应于电话按键上的字母。通过一个映射数组和递归辅助函数实现了这一功能。

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public class Solution {
    public List<String> letterCombinations(String digits) {
        String[] mapping = {"", "", "abc", "def", "ghi", "jkl", "mon", "pqrs", "tuv", "wxyz"};
        List<String> res = new ArrayList<>();
        if (digits == null || digits.length() == 0)
            return res;
        helper(0, digits, mapping, new StringBuilder(), res);
        return res;
    }
    /**
     * @param
     * i, index of current digit
     * digits, digits string
     * mapping, a map between digtis and letters
     * curr, a StringBuilde to save the path from first digit to last
     * res, result list
     */ 
    public void helper(int i, String digits, String[] mapping, StringBuilder curr, List<String> res) {
        if (i == digits.length()) {                         // reache the end of digits, save the path to the result list
            res.add(curr.toString());
            return;
        }
        String str = mapping[digits.charAt(i)-'0'];
        for (int k=0; k<str.length(); k++) {
            curr.append(str.charAt(k));                     // add current lettern to the path
            helper(i+1, digits, mapping, curr, res);        // move to next digit
            curr.setLength(curr.length()-1);                // finish a path, remove the previous letter from the StringBuilder
        }
    }
}