[leetcode]91. Decode Ways解码方法

A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

思路：

解法一：recursion, 找到全部可能的解密方法（会TLE）

time :  O(2 ^ n) 因为是一个二叉树， 其高度最高为n,  所以时间复杂度是O(2 ^ n)

space： O(n)   跟数的高度有关，就是n层

 public int numDecodings(String s ){
     if(s == null || s.length() == 0) retrun 0;
     return numDecodings(s.toCharArray(), 0);
 }

 private int numDecodings(char[] array, int level){
     if(level == array.length){return 1;}
     int ways = 0;
     if(array[level] != ‘0’) {
         ways += numDecodings(array, level + 1);
 　　}
 　　if(validEncoding(array, level)){
     　　ways += numDecodings(array, level + 2);
 　　}
     return ways;
 }

 private boolean validEncoding(char[]array, int start){
     if(start + 1 >= array.length) return false;
     if(array[start] = ‘1’) return true;
     if(array[start] == ‘2’ && array[start + 1] –‘6’ <=0) return true;
     return false;
 }

解法二：自顶向下记忆化搜索（recursion + memorization）

time :  O(1) * (n + 1 )  = O(n)

space： O(n)

 class Solution {
       public int numDecodings(String s ){
         if(s == null || s.length() == 0) return 0;
         int[] m = new int[s.length() + 1];
         Arrays.fill(m, -1);
         return numDecodings(s.toCharArray(), 0, m);
     }

     private int numDecodings(char[] array, int level, int[] m){
         if(m[level] != -1){
             return m[level];
         }
         if(level == array.length){
             m[level] = 1;
             return 1;
         }
         int ways = 0;
         if(array[level] != '0') {
             ways += numDecodings(array, level + 1, m);
         }
         if(validEncoding(array, level)){
             ways += numDecodings(array, level + 2, m);
         }
         return ways;
     }

     private boolean validEncoding(char[]array, int start){
         if(start + 1 >= array.length) return false;
         if(array[start] == '1') return true;
         if( array[start] == '2' && array[start + 1] - '6' <= 0 ){
             return true;
         }
         return false;
     }
 }

解法三： 自底向上的dp

 class Solution {
      public int numDecodings(String s) {
         if (s == null || s.length() == 0) return 0;
         int len = s.length();
         int[] dp = new int[len + 1];
         dp[0] = 1;
         dp[1] = s.charAt(0) == '0' ? 0 : 1;
         for (int i = 2; i <= len; i++) {
             int first = Integer.parseInt(s.substring(i - 1, i));
             int second = Integer.parseInt(s.substring(i - 2, i));
             if (first >= 1 && first <= 9) {
                 dp[i] += dp[i - 1];
             }
             if (second >= 10 && second <= 26) {
                 dp[i] += dp[i - 2];
             }
         }
         return dp[len];
     }
 }