[LeetCode26]Decode Ways

解码字符串的多种可能性：动态规划算法的应用

最新推荐文章于 2025-11-24 01:23:29 发布

原创最新推荐文章于 2025-11-24 01:23:29 发布 · 854 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #decode DP

Leetcode 同时被 2 个专栏收录

287 篇文章

订阅专栏

LeetCode Questions

178 篇文章

订阅专栏

本文深入探讨了如何使用动态规划解决字符串解码问题，具体阐述了解码算法的核心思想、实现步骤及关键条件判断。通过实例演示，展示了如何计算不同编码方式下字符串的解码组合总数。

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.

Analysis:

The key to this problem is : num(current) = num(current-1) + num(current-2).
Be careful with the conditions of above equation:
if the current char =='0'
if the current char and previous char >26
if the previous char =='0'

This can be think as a simple DP problem, res[i]= res[i-1] if s[i] is valid + res[i-2] if s[i-1:i] is valid.

Java

public int numDecodings(String s) {
        if(s.length()<=0) return 0;
        int len = s.length();
        int res[] = new int[len];
        if(s.charAt(0)!='0') res[0] = 1;
        if(len==1) return res[0];
        if(check(s.substring(0, 2))) res[1]++;
        if(s.charAt(0)!='0'&&s.charAt(1)!='0') res[1]++;
        //DP
        for(int i=2;i<len;i++){
        	if(s.charAt(i)!='0')res[i] +=res[i-1];
        	if(check(s.substring(i-1, i+1)))
        		res[i]+=res[i-2];
        }
        return res[len-1];
    }
	public boolean check(String string){
		if(string.length()==0) return false;
		if(string.charAt(0)=='0') return false;
		if(string.length()==2){
			if(string.charAt(0)>'2'||(string.charAt(0)=='2'&&string.charAt(1)>'6'))
				return false;
		}
		return true;
	}

c++

int check(char one){
    return (one != '0') ? 1:0;
}
int check(char one, char two){
    return (one =='1' || (one=='2'&& two<='6'))?1:0;
}
int numDecodings(string s) {
    if(s.empty() || s[0]=='0') return 0;
    if(s.size()==1) return check(s[0]);
    int fn=0, fn_1 = 0,fn_2 = 1;
    fn_1 = (check(s[0])*check(s[1]))+check(s[0],s[1]);
    for(int i=2;i<s.size();i++){
        if(check(s[i])) fn+=fn_1;
        if(check(s[i-1],s[i])) fn+=fn_2;
        if(fn == 0)
            return 0;
        fn_2 = fn_1;
        fn_1 = fn;
        fn = 0;
    }
    return fn_1;
}