1137. N-th Tribonacci Number

The Tribonacci sequence Tn is defined as follows: 

T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.

Given n, return the value of Tn.

 

Example 1:

Input: n = 4Output: 4Explanation:T_3 = 0 + 1 + 1 = 2T_4 = 1 + 1 + 2 = 4

Example 2:

Input: n = 25Output: 1389537

 

Constraints:

• 0 <= n <= 37
• The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.

题目大意：和斐波那契数列差不多，简单。

思路：暴力模拟。

class Solution {    public int tribonacci(int n) {        if( n == 0 || n == 1 ) return n;        if( n == 2 ) return 1;        int sum = 0, three = 1, two = 1, one = 0;        for(int i=3; i<=n; i++) {            sum = three + two + one;            one = two;            two = three;            three = sum;        }        return sum;    }}

 

1138. Alphabet Board Path

On an alphabet board, we start at position (0, 0), corresponding to character board[0][0].

Here, board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"].

We may make the following moves:

• 'U' moves our position up one row, if the square exists;
• 'D' moves our position down one row, if the square exists;
• 'L' moves our position left one column, if the square exists;
• 'R' moves our position right one column, if the square exists;
• '!' adds the character board[r][c] at our current position (r, c) to the answer.

Return a sequence of moves that makes our answer equal to target in the minimum number of moves.  You may return any path that does so.

 

Example 1:

Input: target = "leet"Output: "DDR!UURRR!!DDD!"

Example 2:

Input: target = "code"Output: "RR!DDRR!UUL!R!"

 

Constraints:

• 1 <= target.length <= 100
• target consists only of English lowercase letters.

题目大意：给你一个字母地图board，你最开始是在‘a’的位置，给你一个字符串，让你依次走过每个字母的位置，然后打印路径（随便一个路径就好）。

思路：暴力模拟，但是z有坑，z只能由u下去。

class Solution {    private static int m = 5;    public String alphabetBoardPath(String target) {        StringBuilder res = new StringBuilder();        char[] chars = target.toCharArray();        int len = target.length();        char last = 'a';        for(int i=0; i
      
        last ) {                int ch_o = ch - 'a';                int last_o = last - 'a';                x = ch_o/m - last_o/m;                y = Math.abs(ch_o%m - last_o%m);                for(int k=0; k
       
         last ) {                for(int k=0; k