版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/huatian5/article/details/82795573
题目:https://leetcode.com/problems/decode-ways/description/
题意:让你判断字符串可能有多少解码方式
思路:
当前字符为’0’,则dp[i] = dp[i-2]
当前字符不为’0’,小于等于26,dp[i] = dp[i-1] + dp[2]; 大于26,dp[i] = dp[i-1]
代码:
class Solution {
public:
int numDecodings(string s) {
if(s == "")
return 0;
int dp[s.size()+1] = {1};
for(int i = 1;i <= s.size();i++){
if(s[i-1] != '0')
dp[i] += dp[i-1];
if(i >= 2){
if(s[i-2] == '1' || (s[i-2] == '2' && s[i-1] <= '6'))
dp[i] += dp[i-2];
}
}
return dp[s.size()];
}
};