91. 解码方法

原题

https://leetcode.cn/problems/decode-ways/description/

思路

动态规划

复杂度

时间：O(n)
空间：O(n)

Python代码

class Solution:
    def numDecodings(self, s: str) -> int:
        dp = [0] * len(s)
        for i in range(len(s)):
            if i == 0:
                if '1' <= s[i] <= '9':
                    dp[i] = 1 
                else:
                    dp[i] = 0 
            else:
                if '1' <= s[i] <= '9':
                    dp[i] += dp[i-1]
                if '10' <= s[i-1:i+1] <= '26':
                    if i-2 >= 0:
                        dp[i] += dp[i-2]
                    else:
                        dp[i] += 1
        return dp[-1] 

Go代码

func numDecodings(s string) int {
	dp := make([]int, len(s))
	for i := 0; i < len(s); i++ {
		if i == 0 {
			if '1' <= s[i] && s[i] <= '9' {
				dp[i] = 1
			}
		} else {
			if '1' <= s[i] && s[i] <= '9' {
				dp[i] += dp[i-1]
			}
			n, _ := strconv.Atoi(s[i-1 : i+1])
			if 10 <= n && n <= 26 {
				if i-2 >= 0 {
					dp[i] += dp[i-2]
				} else {
					dp[i] += 1
				}
			}
		}
	}
	return dp[len(dp)-1]
}