leetcode-17 Letter Combinations of a Phone Number

标签：

问题描述：

Given a digitstring, return all possible letter combinations that the number couldrepresent.

A mapping of digitto letters (just like on the telephone buttons) is given below.

Note:

Although the above answer is in lexicographical order, your answer could be inany order you want.

问题分析：简单的树的遍历问题；

代码：

解法一：

/**解法一：直接实现树的遍历*/
public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
		if(digits == null || digits.length() == 0)
			return result;
		
		char[][] chars = {{},//0
						  {},//1
						  {'a','b','c'},//2
						  {'d','e','f'},//3
						  {'g','h','i'},//4
						  {'j','k','l'},//5
						  {'m','n','o'},//6
						  {'p','q','r','s'},//7
						  {'t','u','v'},    //8
						  {'w','x','y','z'} //9
						 };
		char[] temp_str	= new char[digits.length()];
	    DFS(digits, 0, chars, temp_str, result);
		
		return result;				 
	}
	
	/*深度优遍历算法*/
	/*参数：
	  @params:  digits：输入电话号码参数; depth:遍历的树的深度; chars:每个数字对应字符数组;
	            temp_str:临时结果char数组; result:结果List;
	  */
	private void DFS(String digits, int depth, char[][] chars, char[] temp_str, List<String> result)
	{
		//递归的重点
		if(depth == digits.length())
		{
			result.add(new String(temp_str));
			return;
		}
		
		int digit = 0;		
		while((digit = digits.charAt(depth) - '0') < 2)
		{
			depth++;
		}	
		for(int i = 0; i < chars[digit].length; i++)
		{
			//更新temp_str
			temp_str[depth] = chars[digit][i];
			DFS(digits, depth + 1, chars, temp_str, result);
		}	
	}

}

解法二：递归实现
public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
		if(digits == null || digits.length() == 0)
			return result;
		
		char[][] chars = {{},//0
						  {},//1
						  {'a','b','c'},//2
						  {'d','e','f'},//3
						  {'g','h','i'},//4
						  {'j','k','l'},//5
						  {'m','n','o'},//6
						  {'p','q','r','s'},//7
						  {'t','u','v'},    //8
						  {'w','x','y','z'} //9
						 };
		
		//设置每一次循环选取的位置数组
		int[] index = new int[digits.length()];
		
		int k = digits.length() - 1;
		while(k >= 0)
		{
			//临时结果数组
			StringBuffer temp_str = new StringBuffer();
			
			for(int i = 0; i < digits.length(); i++)
			{
				int j = digits.charAt(i) - '0';
				if(j >= 2)
				{
					temp_str.append(chars[j][index[i]]);
				}
			}
			
			result.add(new String(temp_str));
			//每次都从最后一位开始更新
			k = digits.length() - 1;
			//更新index值
			while(k >= 0)
			{
				if(index[k] < (chars[digits.charAt(k)-'0'].length - 1) ) {  
                    index[k]++;  
                    break;  //注意这里的break，当更新完一个值后，就要注意退出
                } 
				else {  
                    index[k] = 0;  
                    k--;  
                }  
			}	
			
		}
		return result;
		
    }
}

leetcode-17 Letter Combinations of a Phone Number

标签：

原文地址：http://blog.csdn.net/woliuyunyicai/article/details/45248613