279. Perfect Squares 类别:medium 难度:medium
2017-01-08 22:21
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题目:
Given a positive integer n, find the least number of perfect square numbers (for example,
For example, given n =
思路:
转移方程:dp[i] = min(dp[i],1 + dp[i - j * j])。
程序:
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1,INT_MAX);
dp[0] = 0;
for(int i = 1;i <= n;i++)
{
for(int j = 1;j * j <= i;j++)
dp[i] = min(dp[i],1 + dp[i - j * j]);
}
return dp
;
}
};
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
3because
12 = 4 + 4 + 4; given n =
13, return
2because
13 = 4 + 9.
思路:
转移方程:dp[i] = min(dp[i],1 + dp[i - j * j])。
程序:
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1,INT_MAX);
dp[0] = 0;
for(int i = 1;i <= n;i++)
{
for(int j = 1;j * j <= i;j++)
dp[i] = min(dp[i],1 + dp[i - j * j]);
}
return dp
;
}
};
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