A message containing letters from A-Z
is being encoded to numbers using the following mapping:
'A' -> 1 'B' -> 2 ... 'Z' -> 26
Given an encoded message containing digits, determine the total number of ways to decode it.
For example,
Given encoded message "12"
, it could be decoded as "AB"
(1 2) or "L"
(12).
The number of ways decoding "12"
is 2.
public class Solution { // dp[i] = '0' < s.char(i) <= '9' ? d[i - 1] : 0 // + 9 < s.sub(i - 2, i) <= 26 ? d[i - 2] : 0 public int numDecodings(String s) { if (s == null || s.length() == 0) return 0; int N = s.length(); int[] dp = new int[N + 1]; dp[0] = 1; // 注意这个初始化条件 dp[1] = (s.charAt(0) >= '1' && s.charAt(0) <= '9') ? 1 : 0; for (int i = 2; i <= N; i++) { char c = s.charAt(i - 1); if (c >= '1' && c <= '9') dp[i] = dp[i - 1]; int prev2 = Integer.valueOf(s.substring(i-2, i)); if (prev2 > 9 && prev2 < 27) dp[i] += dp[i - 2]; } return dp[N]; } /*Time Limit Exceeded O(N^3) public int numDecodings(String s) { if (s == null || s.length() == 0) return 0; int N = s.length(); int[][] dp = new int[N + 1][N + 1]; // initialization dp[i][i+1] for (int i = 0; i < N; i++) { char c = s.charAt(i); dp[i][i + 1] = (c >= '1' && c <= '9') ? 1 : 0; } for (int i = 0; i < N - 1; i++) { String sub = s.substring(i, i + 2); dp[i][i + 2] = (sub.compareTo("10") >= 0 && sub.compareTo("27") < 0) ? 1 : 0; } for (int k = 3; k <= N; k++) { for (int i = 0; i <= N - k ; i++) { for (int j = i + 1; j <= i + k - 1; j++) { dp[i][i + k] += dp[i][j] * dp[j][i + k]; } } } return dp[0][N]; } */ }