17. 电话号码的字母组合

递归

class Solution {
	List<String> ls;
	String[] arr = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

	public List<String> letterCombinations(String digits) {
		ls = new ArrayList<String>();
		ls.clear();
		if (digits != null && digits.length() > 0)
			dfs( digits, "");
		return ls;
	}

	public void dfs( String num, String res) {
		if (num.length() == 0)
			ls.add(res);
		else {
			int flag = num.charAt(0) - '0';
			for (int i = 0; i < arr[flag].length(); i++) {
				dfs( num.substring(1), res + arr[flag].charAt(i));
			}
		}
	}
}

循环

class Solution {

	String[] arr = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

	public List<String> letterCombinations(String digits) {
		List<String> ls = new ArrayList<String>();
		ls.clear();
		if (digits != null && digits.length() > 0) {
            ls.add("");
			for(int i = 0 ; i < digits.length() ;i++) {
				List<String> temp = new ArrayList<String>();
				for(int x = 0 ; x < arr[digits.charAt(i)-'0'].length() ;x++) {
					for(String y : ls) {
						temp.add(y+arr[digits.charAt(i)-'0'].charAt(x));
					}
				}
				ls = temp;
			}
		}
		return ls;
	}
}