279. Perfect Squares
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
.
解:這個問題的解決很類似於 2 Keys Keyboard 這道題,不知道這種類型的動態規劃有沒有特殊的名字,特徵是問題的解由一系列散列的子問題的解可以算得(往往是最小值),拿此題來說,想得到dp[n],可知n可以由n-1加上一個1得到,也可以由n-4加上4得到...,這樣子問題的個數當然就是和n有關的sqrt(n),dp[n] = min{dp[n-1], dp[n-4], ..., dp[n-sqrt(n)*sqrt(n)]}+1,代碼:
class Solution {
public:
int numSquares(int n) {
int *dp = new int[n+1];
dp[1] = 1;
dp[0] = 0;
for (int i = 2; i <= n; i++) {
int temp = sqrt(i);
dp[i] = dp[i-1]+1;
for (int j = 2; j <= temp; j++) {
dp[i] = min(dp[i-j*j] + 1, dp[i]);
}
}
return dp[n];
}
};