LeetCode (D)

Decode Ways

 

AC Rate: 471/2906

My Submissions

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.

class Solution {
public:
	int numDecodings(string s) {
		const int len = s.length();
		if (len == 0) return 0;
		vector<int> dp(len + 1, 0);
		if (s[0] == '0')
			dp[0] = dp[1] = 0;
		else
			dp[0] = dp[1] = 1;
		for (int i = 2; i <= len; ++i) {
			if (s[i - 1] != '0') dp[i] += dp[i - 1];
			const int tmp = (s[i - 2] - '0') * 10 + (s[i - 1] - '0');
			if (tmp >= 10 && tmp <= 26)
				dp[i] += dp[i - 2];
		}
		return dp[len];
	}
};

Divide Two Integers

 

AC Rate: 537/3649

My Submissions

Divide two integers without using multiplication, division and mod operator.

class Solution {
public:
	int divide(int dividend, int divisor) {
		int ans = 0, t = 0;
		long long a = dividend, b = divisor;
		a = a < 0 ? -a : a;
		b = b < 0 ? -b : b;
		while (a >= b) {
			b = b << 1;
			++t;
		}
		b = b >> 1;
		--t;
		while (t >= 0) {
			if (a >= b) {
				a -= b;
				ans += (1 << t);
			}
			--t;
			b = b >> 1;
		}
		return (dividend ^ divisor) < 0 ? -ans : ans;
	}
};

Distinct Subsequences

 

AC Rate: 452/2049

My Submissions

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

class Solution {
public:
	int numDistinct(string S, string T) {
		const int m = S.length(), n = T.length();
		vector<vector<int> > dp(m + 1, vector<int>(n + 1, 0));
		for (int i = 0; i <= m; ++i) dp[i][0] = 1;
		for (int i = 1; i <= m; ++i) {
			for (int j = 1; j <= n; ++j) {
				dp[i][j] = dp[i - 1][j];
				if (S[i - 1] == T[j - 1])
					dp[i][j] += dp[i - 1][j - 1];
			}
		}
		return dp[m][n];
	}
};