【题解】【DP】【Leetcode】Climbing Stairs

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?

思路:

最简单的DP问题:递归+查表

代码:

【题解】【DP】【Leetcode】Climbing Stairs
 1 int climbStairs(int n) {
 2     vector<int> memo(n+1, -1);//忘了在memo[0]没有意义的时候,数组初始大小加一。。。
 3     return climb(n, memo);
 4 }
 5 
 6 int climb(int n, vector<int>& memo){//忘了加&,就会Time Limit Exceeded
 7     if(n < 0){
 8         return -1;
 9     }
10     else if(n <= 2){
11         memo[n] = n;
12         return n;
13     }
14     
15     if (memo[n-1] == -1)
16         memo[n-1] = climb(n-1, memo);//修改了递归函数名之后没有更改其他函数体中的引用
17     if(memo[n-2] == -1)
18         memo[n-2] = climb(n-2, memo);
19     //return memo[n-1] + 2*memo[n-2];//等等,子问题中好像有重叠
20     return memo[n-1] + memo[n-2];
21 }
【题解】【DP】【Leetcode】Climbing Stairs

【题解】【DP】【Leetcode】Climbing Stairs

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