Clock day23.

Topic 1: Swordfinger Offer 19.Regular Expression Matching

Please implement a function to match containment'. 'Regular expressions for and''. Characters in a pattern'.'denote any character, and''denotes that the characters preceding it can occur any number of times (including 0 times). In this topic, matching means that all characters in a string match the entire pattern. For example, the string'a a A' matches the patterns'a.a'and'abaca', but neither'a a.a' nor'ab*a'.

Example 1:
Input:
s = "aa"
p = "a"
Output: false
Interpretation:'a'cannot match the entire string of'a a'.

Example 2:
Input:
s = "aa"
p = "a*"
Output: true
Interpretation: Because'*'means that you can match zero or more of the preceding elements, where the preceding element is'a'. Therefore, the string'a A' can be considered'a'and repeated once.

Example 3:
Input:
s = "ab"
p = "."
Output: true
Interpretation:'. 'means that zero or more ('*') arbitrary characters ('.') can be matched.

Example 4:
Input:
s = "aab"
p = "cab"
Output: true
Interpretation: Because'*'means zero or more, where'c' is zero and'a'is repeated once, the string'aab' can be matched.

Example 5:
Input:
s = "mississippi"
p = "misisp*."
Output: false

s may be empty and contain only lowercase letters from a-z.
p may be empty and contain only lowercase letters from a-z and characters. and, without contiguous''.

Source: LeetCode
Links: https://leetcode-cn.com/problems/zheng-ze-biao-da-shi-pi-pei-lcof

Solving ideas:
Assuming that the main string is A and the pattern string is B, starting from the last step, you need to focus on the last incoming character. Assuming that the length of A is n and B is m, you can focus on who is the last character of the regular expression B. There are three possibilities: normal characters, * and. (point), then you can discuss these three cases.

Big man's puzzle
java code:

class Solution {
    public boolean isMatch(String A, String B) {
        int n = A.length();
        int m = B.length();
        boolean[][] f = new boolean[n + 1][m + 1];

        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                //Divide into empty regular and non-empty regular
                if (j == 0) {
                    f[i][j] = i == 0;
                } else {
                    //Non-empty regular is divided into two cases*and non*
                    if (B.charAt(j - 1) != '*') {
                        if (i > 0 && (A.charAt(i - 1) == B.charAt(j - 1) || B.charAt(j - 1) == '.')) {
                            f[i][j] = f[i - 1][j - 1];
                        }
                    } else {
                        //Encountered*, divided into two situations, see and don't see
                        //Don't look
                        if (j >= 2) {
                            f[i][j] |= f[i][j - 2];
                        }
                        //see
                        if (i >= 1 && j >= 2 && (A.charAt(i - 1) == B.charAt(j - 2) || B.charAt(j - 2) == '.')) {
                            f[i][j] |= f[i - 1][j];
                        }
                    }
                }
            }
        }
        return f[n][m];
    }
}

Question 2: Offer 49. Ugly Number

We call numbers that contain only prime factors 2, 3, and 5 Ugly Numbers. See the nth Ugly Number in order from smallest to largest.

Example:
Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 are the top 10 ugly numbers.

Explain:
1 is an ugly number.
n not more than 1690.

Source: LeetCode
Links: https://leetcode-cn.com/problems/chou-shu-lcof

Solving ideas:

Recursive nature of clowns: clowns contain only factor 2, 3, 52, 3, 5, so there is "clowns==a smaller clowns\times*a factor" (for example, 10 = 5 \times 210=5*2).

java code:

class Solution {
    public int nthUglyNumber(int n) {
        int a = 0, b = 0, c = 0;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i = 1; i < n; i++) {
            int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;
            dp[i] = Math.min(Math.min(n2, n3), n5);
            if(dp[i] == n2) a++;
            if(dp[i] == n3) b++;
            if(dp[i] == n5) c++;
        }
        return dp[n - 1];
    }
}

Question 3: Swordfinger Offer 60.Number of points in n dices

Throw n dices on the floor and sum the number of points all dices face up to s. Enter N and print out the probability of occurrence of all possible values of s.

You need a floating-point array to return the answer, where the ith element represents the probability of the ith smallest of the set of points that this n dice can throw.

Example 1:
Input: 1
Output: [0.16667, 0.16667, 0.16667, 0.16667, 0.16667, 0.16667, 0.16667, 0.16667]

Example 2:
Input: 2
Output: [0.02778, 0.05556, 0.08333, 0.11111, 0.13889, 0.16667, 0.13889, 0.11111, 0.08333, 0.05556, 0.02778]

Restrictions:
1 <= n <= 11

Source: LeetCode
Links: https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof

Solving ideas:
Set the solution of the input n dices (i.e. the list of probabilities) to f(n), where the probability of Points and X is f(n, x). Find the recursive formula.

java code:

class Solution {
    public double[] dicesProbability(int n) {
        double[] dp = new double[6];
        Arrays.fill(dp, 1.0 / 6.0);
        for (int i = 2; i <= n; i++) {
            double[] tmp = new double[5 * i + 1];
            for (int j = 0; j < dp.length; j++) {
                for (int k = 0; k < 6; k++) {
                    tmp[j + k] += dp[j] / 6.0;
                }
            }
            dp = tmp;
        }
        return dp;
    }
}