c++leetcode91解码方法动态规划

解码方法

题解

dp[i]表示到第i-1个字母时解码的方法有多少种

动态转移方程

dp[i] = dp[i-1] dp[i-2]

特殊情况

• s[i]无法和s[i-1]组合：s[i-1]=0 or s[i-1] > 2 or (s[i-1] = 2 && s[i] > 6 ,此时dp[i] = dp[i-1]
• s[i]只能和s[i-1]组合，s[i]=0，此时dp[i] = dp[i-2]

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        if(s[0] == '0')return 0;
        if(n == 1)return 1;
        vector<int> dp(n   1);
        dp[0] = 1;
        dp[1] = 1;

        for(int i = 1;i < n;  i){
            if(s[i] == '0' && (s[i-1] > '2' || s[i-1] == '0'))return 0;
            else if(s[i] == '0')dp[i   1] = dp[i - 1];
            else if(s[i-1] > '2' || s[i-1] == '0' || (s[i-1] >= '2' && s[i] > '6'))
            		dp[i 1] = dp[i];
            else dp[i   1] = dp[i]   dp[i-1];
        }
        return dp[n];
    }
};
//dp[i]=第i位有多少种
//dp[i] = dp[i-1]   dp[i-2]

dp[i] ,dp[i-1],dp[i-2]用c，b，a来表示，节约空间。

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        if(s[0] == '0')return 0;
        if(n == 1)return 1;
        int a = 1,b = 1,c;
        for(int i = 1;i < n;  i){
            if(s[i] == '0' && (s[i-1] > '2' || s[i-1] == '0'))return 0;
            else if(s[i] == '0')c = a;
            else if(s[i-1] > '2' || s[i-1] == '0' || (s[i-1] >= '2' && s[i] > '6'))c = b;
            else c = a   b;
            a = b;
            b = c;
        }
        return c;
    }
};

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