Leetcode Perfect Squares

Problem

Given a positive integer n,
find the least number of perfect square numbers (for example, 

1, 4,
9, 16, ...

) which sum to n.

For example, given n = 

12

, return 

3

 because 

12
= 4 + 4 + 4

; given n = 

13

, return 

2

 because 

13
= 4 + 9

.

Solution 1 - pure recursion

求n 的perfect square, 可以化简成求 n 减去若干个 i*i之后 的数的perfect square.

class Solution {
public:
int numSquares(int n) {
if(n <= 1) return n;

int rst = n;
for( int i = 2; i*i <= n; i++){
int a = n / (i * i), b = n % (i * i);
rst = min (rst, numSquares(b) + a);
}
return rst;
}
};

Solution 2 - recursion with memory

用一个数字，纪录上面的过程中重复值。

class Solution {
int helper( int n, vector<int>& arr){
if(arr
!= -1) return arr
;

int rst = n;
for( int i = 2; i*i <= n; i++){
int a = n / (i * i), b = n % (i * i);
rst = min (rst, helper(b, arr) + a);
}
arr
= rst;
return rst;

}
public:
int numSquares(int n) {
if(n <= 1) return n;
vector<int> arr(n+1, -1);
helper(n, arr);
return arr
;
}
};

Solution 3 - DP

其实上面的解法二已经很接近DP了

dp[ x + y*y ] = min(  dp[ x+ y*y ] , dp[ x ] + 1 );

class Solution {
public:
int numSquares(int n) {
if(n <= 1) return n;
vector<int> dp(n+1, INT_MAX);

for (int i = 1; i * i <= n; i++) {
dp[i * i] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; i + j * j <= n; j++) {
dp[i + j * j] = min(dp[i] + 1, dp[i + j * j]);
}
}
return dp
;
}
};