题目分析
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.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
代码一
时间复杂度为O(2^n),Time Limit Exceeded。
class Solution {
public int climbStairs(int n) {
if(n <= 2) {
return n;
}
return climbStairs(n - 2) + climbStairs(n - 1);
}
}
代码二
动态规划的解法
class Solution {
public int climbStairs(int n) {
if (n == 1) return 1;
int[] num = new int[n + 1];
num[0] = 1;
num[1] = 1;
for (int i = 2; i <= n; i++) {
num[i] = num[i - 1] + num[i - 2];
}
return num[n];
}
}
代码三
class Solution {
public int climbStairs(int n) {
if (n == 1) return 1;
int t1 = 1, t2 = 1, res = 0;
for(int i = 2; i <= n; i++) {
res = t1 + t2;
t2 = t1;
t1 = res;
}
return res;
}
}
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