leetcode279 Perfect Squares

 1 """
 2 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
 3 Example 1:
 4 Input: n = 12
 5 Output: 3
 6 Explanation: 12 = 4 + 4 + 4.
 7 Example 2:
 8 Input: n = 13
 9 Output: 2
10 Explanation: 13 = 4 + 9.
11 """
12 """
13 dp[0] = 0
14 dp[1] = dp[0]+1 = 1
15 dp[2] = dp[1]+1 = 2
16 dp[3] = dp[2]+1 = 3
17 dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 }
18       = Min{ dp[3]+1, dp[0]+1 }
19       = 1
20 dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 }
21       = Min{ dp[4]+1, dp[1]+1 }
22       = 2
23                         .
24                         .
25                         .
26 dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 }
27        = Min{ dp[12]+1, dp[9]+1, dp[4]+1 }
28        = 2
29                         .
30                         .
31                         .
32 dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1
33 is  DP [n] represents the number of least squares and the sum of n is (the request).
34 is  DP all indices array has completely square numbers is a number (e.g. 1,4,9 ...) is set to 1, plus a layer traverse to find the smallest j combined length of the current i.
35  meaning that the dynamic equation: for each i, i is smaller than a full number of those in the square of the minimum number +1 is the desired, i.e. dp [i - j * j]
 + 1. 36  "" " 
37 [  class Solution1:
 38 is      DEF numSquares (Self, n-):
 39          DP = [a float ( ' INF ' )] * (n-+. 1 )
 40          I =. 1
 41 is          the while I * I <= n-:
 42 is              DP [ I * I]. 1 =
 43 is              I = +. 1
 44 is          for I in Range (. 1,. 1 n-+ ):
45             j = 1
46             while j*j <= i:
47                 dp[i] = min(dp[i], dp[i-j*j]+1)
48                 j += 1
49         return dp[n]