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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.
Example 1:
Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
题目：计算组成和为n所需的最少的平方数。
思路：动态规划，同LeetCode322. Coin Change。dp[i]表示组成和为i所需的最少的平方数的个数，通过不同的中间结点j*j，尝试对dp[i]进行松弛。
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n+1, INT_MAX);
dp[0] = 0;
// For each i, it must be the sum of some number (i - j*j)
// and a perfect square number (j*j).
for(int i = 1; i <= n; ++i){
for(int j = 1; j*j <= i; ++j){
dp[i] = min(dp[i], dp[i-j*j]+1);
}
}
return dp[n];
}
};