[Math_Medium] 279. Perfect Squares 2018-09-19
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Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
题目大意:
给定一个数 n, 将其用最少的完全平方数组合出来,比如:12 = 4 + 4 + 4,因此对于输入 12,返回 3; 11 = 9 + 1 + 1,因此也返回 3.
解题思路:
我们可以列举:a[1] = 1;a[2] = a[1] + 1;a[3] = a[1] + a[1] + a[1];
a[4] = 1; a[5] = a[4] + 1; a[6] = a[4] + a[2] ...... a[8] = a[4] + a[4]
a[9] = 1;..........
即可以推导出 a[i] = min(a[i-j*j] + a[j*j])
,其中, j 属于 1 到 sqrt(i)
代码如下:
class Solution {
public:
int numSquares(int n)
{
vector<int> a(n+1, 0);
a[0] = 0;
for(int i = 1; i <= n; i++)
{
int k = sqrt(i);
k = k * k;
if(k == i) //注意: sqrt(2)*sqrt(2) = 2
a[i] = 1;
else
{
int temp = 1<<30 - 1;
for(int j = sqrt(i); j >= 1; j--)
{
if(temp > (a[i-j*j] + a[j*j]))
temp = a[i-j*j] + a[j*j];
}
a[i] = temp;
}
}
return a[n];
}
};
- 推导 2:因为
a[j*j] == 1
, 所以,a[i] = min (a[i], a[i-j*j] + 1)
代码如下:
class Solution {
public:
int numSquares(int n)
{
vector<int> a(n+1, INT_MAX);
a[0] = 0;
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j*j <= i; ++j)
{
a[i] = min(a[i], a[i - j*j] + 1);
}
}
return a[n];
}
};
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