Leetcode---17电话号码的字母组合

思路–暴力

因为digit最长为4，所以直接进行模拟，可以得出最终的答案，时间和空间开销比较大，并且会有很多的

代码–暴力

class Solution {
    String[][] letter = new String[][]{
            {"a","b","c"},
            {"d","e","f"},
            {"g","h","i"},
            {"j","k","l"},
            {"m","n","o"},
            {"p","q","r","s"},
            {"t","u","v"},
            {"w","x","y","z"}
    };
    public List<String> letterCombinations(String digits) {

        List<String> ans = new ArrayList<>();
        int len = digits.length();
        int[] d = new int[4];
        for (int i=0;i<len;i++)
            d[i] = digits.charAt(i)-'0';
        if (d[0]==0)
            return ans;

        for (int i = 0 ;i<letter[d[0]-2].length;i++)
        {
            String cur = letter[d[0]-2][i];
            if (d[1]!=0){
                for (int j = 0 ;j<letter[d[1]-2].length;j++)
                {
                    if (d[2]!=0){
                        for (int k = 0 ;k<letter[d[2]-2].length;k++)
                        {

                            if (d[3]!=0){
                                for (int l = 0 ;l<letter[d[3]-2].length;l++)
                                {
                                    ans.add(cur+letter[d[1]-2][j]+letter[d[2]-2][k]+letter[d[3]-2][l]);
                                }
                            }
                            else
                                ans.add(cur+letter[d[1]-2][j]+letter[d[2]-2][k]);
                        }
                    }
                    else
                        ans.add(cur+letter[d[1]-2][j]);
                }
            }
            else
                ans.add(cur);
        }

        return ans;
    }
}

思路–回溯

思路都是一样的，就是模拟排列组合的过程，只不过这里使用了回溯方法，更加简洁高效。
详情见代码注释。

代码–回溯

class Solution {
    String[][] letter = new String[][]{
            {"a","b","c"},
            {"d","e","f"},
            {"g","h","i"},
            {"j","k","l"},
            {"m","n","o"},
            {"p","q","r","s"},
            {"t","u","v"},
            {"w","x","y","z"}
    };
    List<String> ansLetter = new ArrayList<>();
    StringBuffer sb = new StringBuffer();
    public List<String> letterCombinations(String digits) {

        int len = digits.length();
        if (len==0)//特殊情况判定
            return ansLetter;

        backTrack(digits, 0);//回溯

        return ansLetter;
    }
    public void backTrack(String digits, int index)
    {
        if (sb.length() == digits.length())
        {
            ansLetter.add(sb.toString());
            return;
        }
        //找到对应的字母数组
        String[] val = letter[digits.charAt(index)-'0'-2];
        for (String i : val)
        {
	        //回溯法的基本套路 选择、回溯、退出选择
	        //选择
            sb.append(i);
            //继续向下回溯
            backTrack(digits,index+1);
            //退出选择--这一步的目的是保持StringBuffer的内容不变，以进行下一个元素的回溯
            sb.deleteCharAt(sb.length()-1);
        }
    }
}