[LeetCode-interview algorithm classic-Java implementation] [070-Climbing Stairs (Stairs)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [070-Climbing Stairs (Stairs)], leetcode -- java
[070-Climbing Stairs )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Theme

You are crawling a stair. You need to take n steps to reach the top. You can take two or one step each time. How many different ways can you climb to the top.

Solutions

Solution 1: Use the combination of numbers to solve the problem. In the following cases, C (0, n) does not take two steps at a time, and C (1, n-1), only two steps, C (2, N-2), until only [n/2] (down to take the whole) Walk two steps. The sum is all solutions.
Solution 2: Use the divide and conquer method to save the solution for n steps using an array. a [1] = 1, a [2] = 2, k> = 2, there is a [k] = a [k-1] + a [K-2].

Code Implementation

Algorithm Implementation class, solution 1

Public class Solution {public int climbStairs (int n) {if (n <0) {return 0;} else {int result = 0; for (int I = 0; I <= n; I ++, n --) {result + = combination (I, n);} return result ;}} /*** calculate the number of combinations ** @ param sup superscript * @ param sub subscript * @ return result */private int combination (int sup, int sub) {if (sup> sub | sup <0 | sub <0) {throw new RuntimeException ("Error args");} if (sup = 0) {return 1;} if (sup> sub/2) {sup = sub-sup;} long x = 1; // the product of the denominator long y = 1; // long z; for (int I = 1; I <= sup; I ++) {x * = (sub-I + 1 ); y * = I; z = gcd (x, y); // find the maximum common denominator. // the denominator of a molecule is reduced to x/= z. y/= z ;} return (int) (x/y);} private int gcd (long min, long max) {long tmp; if (max <min) {tmp = min; min = max; max = tmp;} while (max % min! = 0) {tmp = min; min = max % min; max = tmp;} return (int) min ;}}

Algorithm Implementation class, solution 2

Public class Solution {public int climbStairs (int n) {int result = 0; // only the first-order if (n = 1) {result = 1 ;} // only the Level Two else if (n = 2) {result = 2;} // The level Number of the stairs is greater than 2 else if (n> 2) {// save all solutions int [] ways = new int [n]; ways [0] = 1; ways [1] = 2; for (int I = 2; I <ways. length; I ++) {ways [I] = ways [I-1] + ways [I-2];} result = ways [ways. length-1];} return result ;}}

Evaluation Result

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Solution 1

Solution 2

Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/47247995]

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