标签：one make ++ hide none def build lis val
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 = 4 Output: 4 Explanation: T_3 = 0 + 1 + 1 = 2 T_4 = 1 + 1 + 2 = 4
Example 2:
Input: n = 25 Output: 1389537
Constraints:
0 <= n <= 37answer <= 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; } }
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 <= 100target 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<len; i++) { char ch = chars[i]; if( ch == ‘z‘ && ch == last ) { res.append("!"); continue; } if( chars[i] == ‘z‘ ) ch = ‘u‘; int x = 0, y = 0; if( ch > 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<x; k++) { res.append("D"); last = (char) (last + m); } } else if( ch < last ) { int ch_o = ch - ‘a‘; int last_o = last - ‘a‘; x = last_o/m - ch_o/m; y = Math.abs(last_o%m - ch_o%m); for(int k=0; k<x; k++) { res.append("U"); last = (char) (last - m); } } if( ch > last ) { for(int k=0; k<y; k++) res.append("R"); } else if( ch < last ) { for(int k=0; k<y; k++) res.append("L"); } if( chars[i] == ‘z‘ ) res.append("D"); res.append("!"); last = chars[i]; } return res.toString(); } }
标签：one make ++ hide none def build lis val
原文地址：https://www.cnblogs.com/Asimple/p/11258560.html
