91. Decode Ways
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).
Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
解析
输入一个字符串,把它按照规则解析成字母,返回有多少种解析结果。想到要用动态规划,但是想了一会儿没想出来递推的条件,遂放弃。。
参考答案(别人写的)
public class Solution {
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;
dp[1] = s.charAt(0) != '0' ? 1 : 0;
for(int i = 2; i <= n; i++) {
int first = Integer.valueOf(s.substring(i-1, i));
int second = Integer.valueOf(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[n];
}
}
关键点在于递推:
int first = Integer.valueOf(s.substring(i-1, i));
int second = Integer.valueOf(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];
}
仔细推敲这段代码即可。