LeetCode 650. 2 Keys Keyboard--动态规划

题目链接

650. 2 Keys Keyboard

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

Copy All

: You can copy all the characters present on the notepad (partial
copy is not allowed).

Paste

: You can paste the characters which are copied last
time.

Given a number 

n

. You have to get exactly 

n

 'A'
on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get 

n

 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.

解：

dp[i]表示得到i个'A'需要的最少step数，dp[1] = 0，对于dp[i]，他可以由每个能整除i的数j(j > 1)，通过对dp[i/j]做1次Copy All和j-1次Paste得到，而其中能使step最少的一定是所有j中最小的，比如我们想得到dp[24]，

如果通过dp[2]，step总数是2+12 = 14，

如果通过dp[3]，step总数是3+8 = 11，

如果通过dp[4]，step总数则4+6 = 10，

如果通过dp[6]，step总数是5+4 = 9，

如果通过dp[8]，step总数是6+3
= 9，

如果通过dp[12]，step总数是7+2
= 9，

为了减少复杂度，遍历j的范围可以缩小到[2,n/2]，代码：

class Solution {
public:
int minSteps(int n) {
int *dp = new int[n+1];
if (n == 1) return 0;
for (int i = 2; i <= n; i++) {
dp[i] = i;
for (int j = 2; j < n/2+1; j++) {
if (i % j == 0 && (dp[i/j]+j) < dp[i]) {
dp[i] = dp[i/j]+j;
break;
}
}
}
return dp
;
}
};