[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];
    }
}