百度快照搜索引擎,seo查询工具精准,侯马建设规划局网站,wordpress onenote70. 爬楼梯 - 力扣#xff08;LeetCode#xff09;
/** 1阶:1步,即1种; 2阶:1步1步或直接2步,即2种 f(1) 1,f(2) 2 3阶:由1阶迈2步,或2阶迈一步; 4阶:由2阶迈2步,或3阶1步; n阶:由n-2阶迈2步,或n-1阶迈1步 f(n) f(n - 1) f(n - 2)
*/
class Solution {/**1阶:1步,即1种…
70. 爬楼梯 - 力扣LeetCode
/** 1阶:1步,即1种; 2阶:1步1步或直接2步,即2种 f(1) 1,f(2) 2 3阶:由1阶迈2步,或2阶迈一步; 4阶:由2阶迈2步,或3阶1步; n阶:由n-2阶迈2步,或n-1阶迈1步 f(n) f(n - 1) f(n - 2)
*/
class Solution {/**1阶:1步,即1种; 2阶:1步1步或直接2步,即2种f(1) 1,f(2) 23阶:由1阶迈2步,或2阶迈一步; 4阶:由2阶迈2步,或3阶1步; n阶:由n-2阶迈2步,或n-1阶迈1步f(n) f(n - 1) f(n - 2) *///记录已经走过的台阶需要的方法数private ListInteger memory new ArrayList();public int climbStairs(int n) {//初始化,1阶与2阶可直接得出memory.add(1); //f(1) 1;memory.add(2); //f(2) 2;//return help(n);if(n 2) {return memory.get(n - 1);} for(int i 3; i n; i) {memory.add(memory.get(i - 2) memory.get(i - 3));}return memory.get(memory.size() - 1);}private int help(int n) {if(memory.size() n) {return memory.get(n - 1);}if(n 1 || n 2) {return memory.get(n);} else {memory.add(help(n - 1) help(n - 2));}return memory.get(memory.size() - 1);}
}
