leetCode 70. Climbing Stairs
2017-04-04 00:11
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题目如下:
思路:
当n=1,结果为1
当n=2,结果为2
当n=3,结果为3
当n=4,结果为7
易递推
=[n-1]+[n-2],是一个斐波纳切数列。
本题代码:
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? Note: Given n will be a positive integer.
思路:
当n=1,结果为1
当n=2,结果为2
当n=3,结果为3
当n=4,结果为7
易递推
=[n-1]+[n-2],是一个斐波纳切数列。
本题代码:
int climbStairs(int n) { if (n<=0) return 0; if (n==1) return 1; if (n==2) return 2; int one_step_before = 2; int two_steps_before = 1; int all_ways = 0; for(int i=3; i<=n; i++) { all_ways = one_step_before + two_steps_before; two_steps_before = one_step_before; one_step_before = all_ways; } return all_ways; }
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