[leetcode]Decode Ways
动态规划。要注意way+=dp[i-1]或dp[i-2]以及way+=1的条件。我看有的人把数组命名为count的。
public class Solution {
public int numDecodings(String s) {
// Start typing your Java solution below
// DO NOT write main() function
int len = s.length();
if (len == 0) return 0;
int[] dp = new int[len];
for (int i = 0; i < len; i++) {
int way = 0;
if (s.charAt(i) != '0') {
if (i-1 >= 0) {
way += dp[i-1];
}
else {
way += 1;
}
}
if (i-1 >= 0 && (s.charAt(i-1) == '1' || (s.charAt(i-1) == '2' && s.charAt(i) <= '6'))) {
if (i-2 >= 0) {
way += dp[i-2];
}
else {
way+= 1;
}
}
dp[i] = way;
}
return dp[len-1];
}
}
第二刷，做繁琐了
class Solution {
public:
vector<int> ways;
int numDecodings(string s) {
if (s.size() == 0)
return 0;
ways.clear();
ways.resize(s.size() + 1);
for (int i = 0; i < ways.size(); i++)
{
ways[i] = -1;
}
return numDecodingsRe(s, 0);
}
int numDecodingsRe(const string &s, int start)
{
if (ways[start] != -1)
{
return ways[start];
}
if (start == s.size())
{
ways[start] = 1;
return 1;
}
if (s[start] < '1' || s[start] > '9')
{
ways[start] = 0;
return 0;
}
if (start == s.size() - 1)
{
ways[start] = 1;
return 1;
}
if (s[start] == '1')
{
ways[start] = numDecodingsRe(s, start + 1) + numDecodingsRe(s, start + 2);
return ways[start];
}
if (s[start] == '2' && s[start + 1] <= '6')
{
ways[start] = numDecodingsRe(s, start + 1) + numDecodingsRe(s, start + 2);
return ways[start];
}
ways[start] = numDecodingsRe(s, start + 1);
return ways[start];
}
};